convergence of $\sum_{n=1}^\infty \left[\frac{1}{n^s}+\frac{1}{s-1}\left\{\frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}\right\}\right]$ (new methods) The series $\sum_{n=1}^\infty \left[\frac{1}{n^s}+\frac{1}{s-1} \left\{ \frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}\right\}\right]$ is said to converge when $0<s<1,$ which seems impossible, for
$s-1<0,$ $\frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}<0$, and so $$\sum_{n=1}^\infty \left[\frac{1}{n^s}+\frac{1}{s-1} \left\{ \frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}\right\} \right]>\sum_{n=1}^\infty \frac{1}{n^s} \quad (>0),$$ which diverges.
So is the series really convergent?
PS: I would prefer a method without using integral.
And this series is an example of a series whose ratio of nearby items tends to 1.
 A: From Taylor's Theorem, we have
$(1+x)^{t}=1+tx+\frac12 t(t-1)x^2+O(x^3)$.  Appling this result with $x=\frac1n$ and $t=1-s$ reveals that
$$\begin{align}
(n+1)^{1-s}-n^{1-s}&=n^{1-s}\left(\left(1+\frac1n\right)^{1-s}-1\right)\\\\
&=n^{1-s}\left(\frac{1-s}{n}-\frac{\frac12s(1-s)}{n^2}+O\left(\frac1{n^3}\right)\right)\\\\
&=-(s-1)\frac1{n^s}+\frac12s(s-1)\frac1{n^{s+1}}+O\left(\frac1{n^{s+2}}\right)\tag1
\end{align}$$

Using $(1)$, we assert that
$$\frac1{n^s}+\frac1{s-1}\left((n+1)^{1-s}-n^{1-s}\right)=\frac{s/2}{n^{s+1}}+O\left(\frac1{n^{s+2}}\right)$$

Inasmuch as $s\in (0,1)$, the series $\sum_{n=1}^\infty \frac1{n^{1+s}}$ converges and therefore, the series of interest converges also.
A: By the MVT,
$$(n+1)^{1-s}-n^{1-s} = (1-s)c_n^{-s}\cdot 1,$$
where $c_n\in (n,n+1).$ The $n$th term of our series can thus be written
$$n^{-s} +\frac{1}{s-1}(1-s)c_n^{-s} = n^{-s}-c_n^{-s}.$$
Apply MVT again to see this equals $(-s)d_n^{-s-1}(n-c_n),$ where $d_n\in (n,c_n)$ This is no more than
$$s\cdot n^{-s-1}\cdot 1 = \frac{s}{n^{s+1}}.$$
Since $\sum \dfrac{s}{n^{s+1}} <\infty,$ our series converges.
A: In $\sum_{n=1}^\infty \left[\frac{1}{n^s}+\frac{1}{s-1} \left\{ \frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}\right\} \right]$, thinking of the subtraction may result in $\frac{1}{n^\sigma}$ where $\sigma\geq1>s$, I add up the fractions and then get
$$
\frac{u_{n+1}}{u_n}=\frac{1-s-(\frac{n+1}{n+2})^s(n+2)+n+1}{1-s-(\frac{n}{n+1})^s(n+1)+n}(\frac{n}{n+1})^s
=\frac{1-s-(\frac{1}{1+\frac{1}{1+n}})^s(n+2)+n+1}{1-s-(\frac{1}{1+\frac{1}{n}})^s(n+1)+n}(\frac{1}{1+\frac{1}{n}})^s
.$$
Then using $(1+x)^{-s}=1-sx+s(s+1)x^2+O(x^3)$ (for we need to multiply O(n), we need to expand to 2nd order) and changing the denominator to nominator by
$\frac{1}{-s^2+s(s+1)\frac{1}{n}}=-s^2(1-\frac{s(s+1)}{s^2}\frac{1}{n}),$
we have the ratio equals
$$
(-s^2\frac{n}{n+1}-s(s+1)\frac{n}{(n+1)^2})(1-s\frac{1}{n}+s(s+1)\frac{1}{n^2})\frac{(-)}{s^2}(1-(1+\frac{1}{s})\frac{1}{n}).
$$
Then by $\frac{n}{(n+1)^2}=\frac{1}{n(1+\frac{2}{n}+\frac{1}{n^2})}=\frac{1}{n}(1-(\frac{2}{n}+\frac{1}{n^2})+O((\frac{2}{n}+\frac{1}{n^2})^2))=\frac{1}{n}+O(\frac{1}{n^2})$, we have the ratio equals
$$
(1+\frac{1}{s}\frac{1}{n})(1-s\frac{1}{n})(1-(1+\frac{1}{s})\frac{1}{n})+O(\frac{1}{n^2})=1+\frac{-1-s}{n}+O(\frac{1}{n^2}).
$$
Since $-1-s<-1$, we have the series converge absolutely.
(This result also supports the guessing that for $u_n=1/n^\lambda$, $\frac{u_{n+1}}{u_n}$ equals $1-\lambda/n$.
My thought is to try using the ratio test. But even if so, I can try first to simplify $u_n$ to $\alpha_0+\frac{\alpha_1}{n}+O(\frac{1}{n^2}) $ (or 2nd order expansion, etc.) before I do the division, which will make the calculation easier. But the above calculation also verify however and in whatever order we do the calculation, the expansion of a function (here, of 1/n) won't change. Such perfect consistence is so nice, considering how different the paths the result is gotten via.
