Does the Series $\sum \frac{e^n}{n^2}$ Converge or Diverge?

Q: Does the series sum e to the n over n squared converge?

No. The ratio test gives a limit of e, which is greater than 1, so the series diverges.

Q: Why does an exponential numerator cause divergence here?

Because exponential growth eventually outpaces any polynomial denominator such as n squared, so the terms do not shrink fast enough for the series to converge.

Question

13) $$\sum_{n=1}^{\infty} \frac{e^n}{n^2}$$
$$a_n = \frac{e^n}{n^2} \qquad a_{n+1} = \frac{e^{n+1}}{(n+1)^2}$$
$$\lim_{n	o\infty} \left| \frac{e^n}{n^2} \cdot \frac{(n+1)^2}{e^{n+1}} ight|$$
$$\lim_{n	o\infty} \left| \frac{n^2 + 2n + 1}{e n^2} ight| = \frac{1}{e} < 1$$
So it converges because of ratio test

Acemy AI · Accepted Answer

**Key takeaway**: This one looks harmless at first, but the exponential factor is the real story. Once $e^n$ appears in the numerator, polynomial denominators usually cannot keep up.

### Step 1: Write the terms
Let

$$a_n=\frac{e^n}{n^2}.$$

### Step 2: Compute the ratio

$$\left|\frac{a_{n+1}}{a_n}\right|
=\frac{e^{n+1}}{(n+1)^2}\cdot\frac{n^2}{e^n}
=e\cdot \frac{n^2}{(n+1)^2}.$$

Now take the limit:

$$\lim_{n\to\infty} e\cdot \frac{n^2}{(n+1)^2}=e.$$

### Step 3: Interpret the result
Because

$$e>1,$$

the ratio test says the series

$$\sum_{n=1}^{\infty}\frac{e^n}{n^2}$$

diverges.

### Why the earlier limit $1/e<1$ is misleading
If you accidentally invert the ratio, you may get $1/e$. For the ratio test, the limit has to be $\left|a_{n+1}/a_n\right|$, not $\left|a_n/a_{n+1}\right|$. That distinction changes the conclusion completely.

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**Pitfalls the pros know** 👇
The main trap here is using the ratio in the wrong direction. For ratio test work, always form $a_{n+1}/a_n$. Also, a denominator like $n^2$ cannot offset an exponential numerator like $e^n$.

**What if the problem changes?**
If the series were $\sum \frac{e^{-n}}{n^2}$, then the ratio would be $\frac{1}{e}\cdot\frac{n^2}{(n+1)^2}<1$, so it would converge. If the numerator were $2^n$ instead of $e^n$, the same ratio-test structure would apply, with limit 2.

`Tags`: ratio test, exponential growth, divergent series

Question

Does the Series \(\sum \frac{e^n}{n^2}\) Converge or Diverge?

Expert Verified Solution

Step 1: Write the terms

Step 2: Compute the ratio

FAQ

Does the series sum e to the n over n squared converge?

Why does an exponential numerator cause divergence here?