Convergent and Divergent Sequences: Tests and Examples

What Are Convergent And Divergent Sequences?

Convergent and divergent sequences are the two possible outcomes when you follow an infinite list of numbers to its end. A sequence ${a_n}$ is convergent if its terms get and stay as close as you like to one finite number $L$, called the limit:

$$\lim_{n \to \infty} a_n = L.$$

A sequence is divergent if no such finite $L$ exists. That is the whole definition, and it answers the question "What are convergent and divergent sequences?" in two lines. The precise epsilon-N version of "as close as you like" is worked through in limit of a sequence; this page is about the practical job of deciding which case you are in.

Three facts frame everything below:

• The limit is unique. A sequence cannot converge to two different numbers.

• Only the tail matters. Changing the first thousand terms never changes whether a sequence converges, or its limit.

• Divergent is not one thing. A sequence can diverge by growing without bound, by falling without bound, or by refusing to settle.

What Are The Five Ways A Sequence Can Behave?

Every real sequence lands in exactly one of five boxes. Many textbooks (especially in India) use the terms oscillates finitely and oscillates infinitely for the last two.

Writing $\lim_{n \to \infty} n^2 = \infty$ is a convenient shorthand, not a limit in the convergent sense. The sequence still diverges; the notation simply records how.

When Can You Say A Sequence Is Convergent Or Divergent?

You can say a sequence converges once you have found its limit with a valid method, or proved that a limit exists. You can say it diverges once you have shown it grows without bound, or found two parts of it heading to different places. The question "When can you say that the sequence is convergent or divergent?" usually comes from students who tried listing terms, which can suggest an answer but never proves one.

A reliable order of attack:

1. Look for a known form. Geometric $r^n$, a rational expression in $n$, or a standard limit such as $\frac{1}{n} \to 0$.

2. Use algebra and the limit laws. Divide by the highest power of $n$, rationalise, or split into simpler pieces.

3. Pass to a function. If $a_n = f(n)$ and $f(x)$ has a limit as $x \to \infty$, use it, with L'Hôpital's rule when the form is indeterminate.

4. Trap it. Bound $a_n$ between two sequences with the same limit (squeeze).

5. Split it. If even and odd terms head to different limits, the sequence diverges.

6. Fall back on structure. If none of the above gives a limit, show the sequence is monotonic and bounded, which forces convergence (covered in monotonic sequences).

How Do You Find The Limit Of A Sequence Using Algebra?

Most textbook sequences surrender to algebra. The limit laws work for sequences exactly as for functions: if $a_n \to A$ and $b_n \to B$, then $a_n + b_n \to A + B$, $a_n b_n \to AB$, and $\frac{a_n}{b_n} \to \frac{A}{B}$ when $B \neq 0$.

For a rational expression, divide top and bottom by the highest power of $n$ in the denominator. Every term with a leftover $\frac{1}{n^k}$ vanishes, so only the leading terms matter:

When the expression is a difference of roots, such as $\sqrt{n^2+n} - n$, the limit laws do not apply directly, because $\infty - \infty$ is one of the indeterminate forms. Multiply by the conjugate first (Mistake 1 below shows why this matters).

How Does L'Hôpital's Rule Help With Sequences?

L'Hôpital's rule works on functions of a continuous variable, not on sequences, so you borrow a function first. The bridge is a theorem:

$$\text{If } a_n = f(n) \text{ and } \lim_{x \to \infty} f(x) = L, \text{ then } \lim_{n \to \infty} a_n = L.$$

Here $f$ is a function defined for real $x \geq 1$ whose values at the whole numbers are the terms of the sequence. Once you are working with $f(x)$, L'Hôpital's rule is available for $\frac{\infty}{\infty}$ and $\frac{0}{0}$ forms: differentiate the top and bottom separately and take the limit again.

The theorem only runs one way. The sequence $a_n = \sin(\pi n)$ equals $0$ for every whole number $n$, so it converges to $0$, yet $f(x) = \sin(\pi x)$ keeps oscillating between $-1$ and $1$ and has no limit. A function with no limit says nothing about the sequence; you have to check the sequence another way. (I find this the least intuitive point in the topic: the sequence only ever looks at the function on whole numbers, and those can line up with the zeros of a wave.)

How Does The Squeeze Theorem Decide Convergence?

The squeeze theorem settles sequences that wobble but shrink. If

$$b_n \leq a_n \leq c_n \quad \text{for all large } n, \qquad \text{and} \qquad b_n \to L, ; c_n \to L,$$

then $a_n \to L$ as well. The standard use is a bounded factor times something that goes to zero, for example $\frac{\cos n}{n}$, which is trapped between $-\frac{1}{n}$ and $\frac{1}{n}$. The function version, with the classic trigonometric limits, is in squeeze theorem.

A useful special case is the absolute value theorem:

$$\text{if } |a_n| \to 0, \text{ then } a_n \to 0.$$

It works only for the limit $0$. Knowing $|a_n| \to 1$ tells you nothing about $a_n$ itself, as Example 2 shows.

What Happens To The Geometric Sequence $r^n$?

The powers $r^n$ appear inside so many problems that their behaviour is worth knowing by heart. A geometric sequence multiplies by the same ratio $r$ each step, and the size of $r$ decides everything:

So $r^n$ converges exactly when $-1 < r \leq 1$. The negative ratios between $-1$ and $0$ still converge, because the sign flips but the size shrinks: $(-0.8)^n$ runs $-0.8, 0.64, -0.512, \ldots$ toward $0$.

How Do Subsequences Prove That A Sequence Diverges?

A subsequence keeps infinitely many terms in their original order, such as all the even-numbered terms. If a sequence converges to $L$, every subsequence converges to the same $L$. Turn that around and you get the most useful divergence test there is:

$$\text{If two subsequences have different limits, the sequence diverges.}$$

For $(-1)^n$, the even terms are all $1$ and the odd terms are all $-1$, so it diverges.

The question "Does every bounded, divergent sequence contain only convergent subsequences with at least two different limits?" deserves a careful answer. Not every subsequence of a bounded divergent sequence converges (the sequence itself is one of its subsequences, and it diverges). What is true is this: a bounded sequence that diverges always has at least two convergent subsequences with different limits. For $(-1)^n \frac{n}{n+1}$ those limits are $1$ and $-1$.

How Are Convergent Sequences Different From Convergent Series?

A sequence is a list; a series is the sum of that list. The question "What is the difference between a convergent series and a divergent series in Calculus II?" often hides a mix-up between the two, so keep them separate:

• Sequence: $a_1, a_2, a_3, \ldots$ converges if the terms themselves settle.

• Series: $a_1 + a_2 + a_3 + \cdots$ converges if the sequence of running totals (the partial sums) settles.

The harmonic example from the hook shows the gap. The terms $\frac{1}{n}$ converge to $0$, but the partial sums do not settle:

$$1 + \frac{1}{2} + \cdots + \frac{1}{100} \approx 5.1874, \qquad 1 + \frac{1}{2} + \cdots + \frac{1}{10^6} \approx 14.3927.$$

They grow like $\ln n$, slowly but without bound, which is why the harmonic series diverges. The one firm link between the two ideas runs in a single direction: if the terms do not converge to $0$, the series must diverge. That is the divergence test, and it can only ever prove divergence. Everything else about summing is in convergence and divergence of series.

Examples Of Convergent And Divergent Sequences

Example 1

Decide whether $a_n = \frac{3n^2 - 5n}{4n^2 + 1}$ converges, and find its limit.

Divide top and bottom by $n^2$:

$$a_n = \frac{3 - \frac{5}{n}}{4 + \frac{1}{n^2}}.$$

As $n \to \infty$, $\frac{5}{n} \to 0$ and $\frac{1}{n^2} \to 0$:

$$\lim_{n \to \infty} a_n = \frac{3 - 0}{4 + 0} = \frac{3}{4}.$$

Final answer: converges to $\frac{3}{4}$.

Example 2

Decide whether $a_n = (-1)^n \frac{n}{n+1}$ converges.

A quick attempt ignores the sign factor and takes the limit of the fraction:

$$\lim_{n \to \infty} \frac{n}{n+1} = 1, \quad \text{so } a_n \to 1 \quad (\text{wrong}).$$

Listing terms breaks this at once:

$$a_1 = -\frac{1}{2}, \quad a_2 = \frac{2}{3}, \quad a_3 = -\frac{3}{4}, \quad a_4 = \frac{4}{5}.$$

Half the terms are negative and moving toward $-1$, so they cannot be settling near $1$. Split by parity. For even $n$:

$$a_{2k} = \frac{2k}{2k+1} \to 1.$$

For odd $n$:

$$a_{2k-1} = -\frac{2k-1}{2k} \to -1.$$

Two subsequences, two different limits. The size $|a_n|$ does tend to $1$, but the absolute value theorem only works for a limit of $0$.

Final answer: diverges (oscillates finitely between values near $1$ and $-1$).

Example 3

Find $\lim_{n \to \infty} \frac{(\ln n)^2}{n}$.

Pass to the function $f(x) = \frac{(\ln x)^2}{x}$, which has the form $\frac{\infty}{\infty}$. Apply L'Hôpital's rule once:

$$\lim_{x \to \infty} \frac{(\ln x)^2}{x} = \lim_{x \to \infty} \frac{\frac{2 \ln x}{x}}{1} = \lim_{x \to \infty} \frac{2 \ln x}{x}.$$

Still $\frac{\infty}{\infty}$, so apply it again:

$$\lim_{x \to \infty} \frac{2 \ln x}{x} = \lim_{x \to \infty} \frac{\frac{2}{x}}{1} = 0.$$

The function has limit $0$, so the sequence does too.

Final answer: converges to $0$ (any power of $\ln n$ loses to $n$).

Example 4

Find $\lim_{n \to \infty} \frac{3n + \sin n}{n}$.

Split the fraction:

$$a_n = 3 + \frac{\sin n}{n}.$$

Since $-1 \leq \sin n \leq 1$:

$$-\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}.$$

Both outer sequences tend to $0$, so by the squeeze theorem $\frac{\sin n}{n} \to 0$, and by the sum law:

$$\lim_{n \to \infty} a_n = 3 + 0 = 3.$$

Final answer: converges to $3$.

Example 5

Find $\lim_{n \to \infty} \left(1 - \frac{2}{n}\right)^n$.

The form is $1^\infty$, which is indeterminate. Take the natural logarithm of the function version:

$$\ln y = x \ln\left(1 - \frac{2}{x}\right) = \frac{\ln\left(1 - \frac{2}{x}\right)}{\frac{1}{x}}.$$

This is $\frac{0}{0}$. By L'Hôpital's rule:

$$\lim_{x \to \infty} \frac{\frac{2}{x^2} \cdot \frac{1}{1 - \frac{2}{x}}}{-\frac{1}{x^2}} = \lim_{x \to \infty} \frac{-2}{1 - \frac{2}{x}} = -2.$$

So $\ln y \to -2$ and

$$\lim_{n \to \infty} \left(1 - \frac{2}{n}\right)^n = e^{-2} \approx 0.1353.$$

Final answer: converges to $e^{-2}$.

Example 6

A ball is dropped from 2 m and each bounce reaches 75% of the previous height. Does the sequence of bounce heights converge, and is the total distance travelled finite?

The height after the $n$th bounce is

$$h_n = 2,(0.75)^n ,\text{m}: \quad 1.5, ; 1.125, ; 0.84375, \ldots$$

This is geometric with $r = 0.75$, and $-1 < r < 1$, so

$$\lim_{n \to \infty} h_n = 0.$$

The total distance is a separate question about a series: the first 2 m fall, then each bounce height travelled up and down:

$$D = 2 + 2\sum_{n=1}^{\infty} 2,(0.75)^n = 2 + 2 \cdot \frac{1.5}{1 - 0.75} = 2 + 12 = 14 ,\text{m}.$$

Final answer: the heights converge to $0$; the total distance is finite, 14 m. Here both the sequence and the series converge, unlike the harmonic case.

Why Does It Matter Whether A Sequence Converges?

The short reason: "a computer can only give you the sequence, never the limit." Square roots, logarithms, $\pi$ and the solutions of most equations are computed by producing better and better approximations, one term at a time. Convergence is the guarantee that those terms are honing in on something real, and divergence is the warning that they are not.

• Iterative algorithms. Newton's method generates a sequence of guesses for a root. If the sequence converges, the root is found to any accuracy you want; if it oscillates or blows up, more computing time makes things worse, not better.

• Defining numbers. Irrational numbers are pinned down as limits of sequences of rationals, and $e$ itself is the limit of $\left(1 + \frac{1}{n}\right)^n$.

• Modelling. Repeated processes, from loan balances to drug levels to population counts, produce sequences. Whether they settle, explode or oscillate is often the only question the model has to answer.

Where Are Convergent And Divergent Sequences Used In Real Life?

• Finance: a balance growing by a fixed percentage, $P(1.05)^n$, diverges to infinity, while repeated repayments can drive a debt sequence to $0$.

• Physics: bounce heights and damped vibrations shrink geometrically, so their sequences converge to $0$.

• Computing: numerical solvers stop when successive terms differ by less than a tolerance, a practical stand-in for convergence.

• Signal processing: sampled signals are sequences, and an echo or feedback loop whose samples grow instead of fading is a divergent sequence heard as a howl.

Convergent And Divergent Sequences: Where Intuition Breaks

Mistake 1: Treating $\infty - \infty$ As Zero

Where it slips in: Limits like $\sqrt{n^2 + n} - n$, where both parts grow without bound.

Don't do this: Write "both terms go to infinity, so the difference goes to $0$."

The correct way: Multiply by the conjugate:

$$\sqrt{n^2 + n} - n = \frac{n}{\sqrt{n^2 + n} + n} = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} \to \frac{1}{2}.$$

The rusher reaches for "infinity minus infinity is zero" because the two pieces look alike. The habit that fixes it is naming the form before computing: once you write "$\infty - \infty$, indeterminate", the conjugate step follows naturally.

Mistake 2: Reading Divergence From The Function

Where it slips in: Using the function theorem in reverse, for example concluding that $\sin(\pi n)$ diverges because $\sin(\pi x)$ has no limit.

Don't do this: Assume a function with no limit means the sequence has no limit.

The correct way: Use the function theorem only when the function has a limit. When it does not, test the sequence directly: here every term is $0$, so the sequence converges to $0$.

Mistake 3: Thinking Terms Going To Zero Make A Sum Finite

Where it slips in: Moving from sequences to series, for instance assuming $1 + \frac{1}{2} + \frac{1}{3} + \cdots$ must be finite because $\frac{1}{n} \to 0$.

Don't do this: Treat "terms converge to $0$" as proof that the series converges.

The correct way: Treat $a_n \to 0$ as a necessary condition only. It rules nothing in, and the harmonic series is the standard counterexample.

The memorizer runs into this one. A student who has memorised "the terms must go to zero" often hears it as "if the terms go to zero, the series converges", which reverses the rule. Writing the divergence test as "terms not going to 0 means divergence" keeps the direction straight.

Practice Problems On Convergent And Divergent Sequences

1. Find $\lim_{n \to \infty} \frac{5n^3 - 2}{2n^3 + n}$.

2. Decide whether $a_n = \frac{n}{\ln n}$ converges.

3. Decide whether $a_n = \frac{\cos n}{\sqrt{n}}$ converges.

4. Decide whether $a_n = (-1)^n \left(2 + \frac{1}{n}\right)$ converges.

5. Find $\lim_{n \to \infty} \left(1 + \frac{1}{2n}\right)^n$.

6. Find $\lim_{n \to \infty} \left(\sqrt{n^2 + 4n} - n\right)$.

Answer to Question 1: Equal degrees, so the limit is the ratio of leading coefficients, $\frac{5}{2}$.

Answer to Question 2: By L'Hôpital on $\frac{x}{\ln x}$, the limit is $\lim_{x \to \infty} \frac{1}{1/x} = \infty$, so the sequence diverges to $+\infty$.

Answer to Question 3: $\left|\frac{\cos n}{\sqrt{n}}\right| \leq \frac{1}{\sqrt{n}} \to 0$, so it converges to $0$ by the squeeze theorem.

Answer to Question 4: The terms are $-3, \frac{5}{2}, -\frac{7}{3}, \frac{9}{4}, \ldots$; even terms tend to $2$ and odd terms to $-2$, so it diverges (oscillates finitely).

Answer to Question 5: The logarithm form gives $\ln y \to \frac{1}{2}$, so the limit is $e^{1/2} \approx 1.6487$.

Answer to Question 6: Multiply by the conjugate to get $\frac{4n}{\sqrt{n^2 + 4n} + n} \to \frac{4}{2} = 2$.

Key Takeaways

• Convergent and divergent sequences differ in one way: a convergent sequence settles on a single finite limit, and a divergent one does not.

• A sequence can diverge to $+\infty$, to $-\infty$, or by oscillating finitely or infinitely.

• Limit laws, L'Hôpital through a function, the squeeze theorem and the geometric rule for $r^n$ settle most sequences.

• Two subsequences with different limits prove divergence.

• Terms converging to $0$ never prove that a series converges.

Where To Go From Here

Work through Practice Problems 1 to 6 in order; if Problem 4 trips you up, return to Example 2 and its parity split, and if Problem 5 does, redo the logarithm steps in Example 5 line by line. Want a live Bhanzu trainer to walk through more convergence problems? Book a free demo class.

Read More

• Infinite series and their partial sums

• Geometric series and their sums

• Limits at infinity for functions

• Standard limits worth memorising

• The natural logarithm and its properties