# Convergence of series through asymptote

Going through an exercise and trying to find the convergence of this series: $$\sum_{n=1}^N \left(e^\frac{1}{n}-1 - \frac{1}{n}\right)$$ The asymptotic of $(e^\frac{1}{n}-1)$ is $\frac{1}{n}$ as n tends to Infinity, so it is for the remaining part, this leads to say that the series does not converge. But in the explanation, it makes an asymptotic of the series as $$(e^\frac{1}{n}-1 - \frac{1}{n})\sim\frac{1}{2n^2}$$ Thus, it seems that the series converges. Can someone explain why separating the function is wrong and how they found the asymptote of the function?

• It is not wrong to start off by considering the combination $e^{\frac{1}{n}}-1$, however when you find that this is $\sim \frac{1}{n}$ which is cancelled by the term $-\frac{1}{n}$ you left out then you need to go back and consider the next to leading order term in this expansion to find the correct asymptotic behavior of the summand. In any case you cannot ignore the $-\frac{1}{n}$ term which you seem to do with your first argument. – Winther Feb 13 '17 at 0:01
• So you are saying that everytime i find an indefinite form i need to continue with expansion until i find the convergence or divergence of the series? – Harry Feb 13 '17 at 0:07
• I'm saying that you need to consider all the terms in the summand to be able to conclude. If we have the series $e^{\frac{1}{n}} - 1 - \frac{1}{n^2}$ then the estimate $e^{\frac{1}{n}} - 1 \sim \frac{1}{n}$ is enough as this is much larger than $-\frac{1}{n^2}$ so this will be the dominant term and the series will diverge. In your case $\frac{1}{n}$ is cancelled by $-\frac{1}{n}$ so in this case you should go to the next order. You need to expand until you have found the leading order term. – Winther Feb 13 '17 at 0:10
• Got it :) thanks for the help! – Harry Feb 13 '17 at 0:13

The first approximations to $e^x$ as $x \to 0$ are $e^x \approx 1+O(x)$, $e^x \approx 1+x+O(x^2)$, and $e^x \approx 1+x+x^2/2+O(x^3)$.

Putting these in $s(N) =\sum_{n=1}^N \left(e^\frac{1}{n}-1 - \frac{1}{n}\right)$, we first get $s(N) =\sum_{n=1}^N \left(1+O(1/n)-1 - \frac{1}{n}\right) =\sum_{n=1}^N \left(O(1/n)\right)$, which does not tell us about convergence.

For the next approximation, $s(N) =\sum_{n=1}^N \left(1+1/n+O(1/n^2)-1 - \frac{1}{n}\right) =\sum_{n=1}^N \left(O(1/n^2)\right)$, which tells us, if the coefficient implied by the $O(1/n^2)$ is bounded, that the sum converges.

Since the coefficient is bounded, the sum does converge and there is no need to go to the next approximation.

• Very well, explained! :) – Harry Feb 13 '17 at 0:05

$e^{1/n}-1-\frac{1}{n}$ behaves like $\frac{1}{2n^2}$ for large $n$s hence the given series is convergent by the p-test.
Additionally, $$\sum_{n\geq 1}\left(e^{1/n}-1-\frac{1}{n}\right) = \sum_{k\geq 2}\frac{1}{k!}\sum_{n\geq 1}\frac{1}{n^k} = \sum_{k\geq 2}\frac{\zeta(k)}{k!}$$ can be represented as the fast-converging integral $2\int_{0}^{+\infty}\frac{-x+ I_1(2x)}{e^{x^2}-1}\,dx.$

Hint: $$e^{1/n} = 1 + \frac{1}{n} + \frac{1}{2n^2} + O(\frac{1}{n^3})$$