# Determine whether the series $\frac{e^\frac{1}{n}}{n^2}$ converges or diverges

The problem states

Determine whether the series converges or diverges: $$\sum_{n=1}^\infty\frac{e^{1/n}}{n^2}$$

Immediately I always like to try the test for divergence:

$$\lim_{n\to\infty} \sum_{n=1}^\infty\frac{e^{1/n}}{n^2} = 0$$ I judge this to be true by looking at the numerator and denominator of the series. The numerator will tend to 1 as $$n\to \infty$$ and the denominator will grow bigger and bigger, so eventually you will have $$\frac{1}{\infty}$$ which will equal to zero. This tells us absolutely nothing about the convergence or divergence of this series.

Next I'll try the integral test because i see in the denominator a term resulting from the derivative of the numerator. To apply the integral test the sequence $$a_n$$ of our series $$\sum_{n=1}^\infty a_n$$ should be continuous, positive and decreasing. The summation begins at one and not zero so we have no issues with continuity here. Both numerator and denominator will yield positive terms as $$n \to \infty$$ and sequential terms will get smaller as the numerator shrinks whilst the denominator grows as $$n\to \infty$$.

$$\text{let } f(x) = \frac{e^{x^{-1}}}{x^2}$$ $$\int_1^\infty f(x) dx = \int_1^\infty \frac{e^{x^{-1}}}{x^2}dx$$ $$= -e^{x{^{-1}}}|_1^\infty$$ $$= \lim_{x\to \infty}- e^{\frac{1}{x}} +e$$ $$=-1 + e$$

because $$\int_1^\infty f(x)$$ converges then by the integral test so must our series.

I have two questions:

1. How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?
2. Would it be possible to determine whether this series converges or diverges using the limit comparison test?
• Compare, to say $e/n^2$. – Angina Seng Oct 1 '17 at 16:19
• This makes no sense: $$\lim_{n\to\infty} \sum_{n=1}^\infty\frac{e^{\frac{1}{n}}}{n^2} = 0$$ – Simply Beautiful Art Oct 1 '17 at 16:27
• The test for divergence is applied to the sequence and not the series? I.e) if i had written $$\lim_{n\to\infty} \frac{e^{\frac{1}{n}}}{n^2} = 0$$ that would be fine? – Blargian Oct 1 '17 at 16:49
• Yes, that is correct. – Simply Beautiful Art Oct 1 '17 at 16:53
• Don’t forget the $dx$ on your integral – gen-ℤ ready to perish Oct 1 '17 at 18:24

As Lord Shark the Unknown comments, simply note that

$$e^{1/n}<e,\quad n>1$$

And so,

$$0<\sum_{n=1}^\infty\frac{e^{1/n}}{n^2}<\sum_{n=1}^\infty\frac e{n^2}=\frac{e\pi^2}6<\infty$$

How can i write the conditions for the integral test (continuous,positive,decreasing) in a mathematical way. What would be the mathematically accepted way to show these conditions?

If $f(x)$ is continuous, $f(x)\ge0$, and $f'(x)\le0$, then...

Would it be possible to determine whether this series converges or diverges using the limit comparison test?

Sure. You could apply the limit comparison test against $1/n^2$, since $e^{1/n}\to1$ as $n\to\infty$.

• Could you please explain how $$\sum_{n=1}^\infty\frac e{n^2}=\frac{e\pi^2}6$$ – Blargian Oct 1 '17 at 16:50
• – Simply Beautiful Art Oct 1 '17 at 16:52
• You probably don't want this, but another interpretation of the conditions for the integral test would be$$(f\in C^1)\land\forall x\ge0[(f(x)\ge0)\land f'(x)\le0)]\implies\left(\sum_{n=0}^\infty f(n)<\infty\iff\int_0^\infty f(x)~\mathrm dx<\infty\right)$$ – Simply Beautiful Art Oct 1 '17 at 16:56
• If you don't want to restrict this to differentiable functions, then use $\forall x\ge0\forall y\ge0[x<y\implies f(x)>f(y)]$ – Simply Beautiful Art Oct 1 '17 at 16:57