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