Convergence of (-1)^n * sqrt(n)/n series What I am having problems with is the following series.
$$\sum_{n=1}^\infty (-1)^n\frac{\sqrt{n}}{n+907}$$
So far I have tried Abel's test and while It is apparent that the series $$(-1)^n \frac{1}{n+907}$$ converges, I couldn't prove that $\sqrt{n}$ is bounded above.
The next thing I tried was Leibniz test for alternating series and while I could prove that
$$\lim_{n \to\infty}\frac{1}{n+907}=0$$ I couldn't prove that $\frac{1}{n+907}$ is monotonically decreasing.
Up to now I have run out of ideas about what to try. Note that I cannot test for absolute convergence, I must do it by another method.
 A: If you don't want to prove that monotonicity (or in cases where it is difficult to prove it), a small expansion can be used :
$$(-1)^n\frac{\sqrt{n}}{n+907} =\frac{(-1)^n}{\sqrt{n}}\left(\frac{1}{1+\frac{907}{n}} \right)=\frac{(-1)^n}{\sqrt{n}}\times\left(1-\frac{907}{n} + o \left(\frac{1}{n} \right) \right)$$ so $$ (-1)^n\frac{\sqrt{n}}{n+907} =\frac{(-1)^n}{\sqrt{n}}-\frac{907(-1)^n}{n\sqrt{n}} + o \left(\frac{1}{n\sqrt{n}} \right) $$
and all the terms here are the terms of convergent series, so the initial series converges.
A: The function $\frac{\sqrt x}{x+907}$ is monotonically decreasing – the numerator of derivative (the denominator is always positive, so we can ignore it) is
$$(x+907)/(2\sqrt x)-\sqrt x=\frac{907}{2\sqrt x}-\frac12\sqrt x$$
and since $\sqrt x$ grows faster than $1/\sqrt x$, this derivative is negative at sufficiently large $x$. Thus, by the alternating series test, the series converges.
A: $\sqrt n$ is evidently not bounded, therefore trying to prove that it is bounded is a waste of time.
$\frac1{n+907}$ is monotonically decreasing in quite an easy way, however that fact is completely irrelevant for the purpose of using Leibniz criterion: that criterion states quite clearly that $\sum_{k=0}^\infty (-1)^k a_k$ converges if $a_k$ decreases to $0$.
However, what is true is that $\frac{\sqrt n}{n+907}$ is positive, it converges to $0$ and it is eventually decreasing, meaning that there is some $M$ such that, for all $n>m>M$, $\frac{\sqrt m}{m+907}>\frac{\sqrt n}{n+907}$. Since you have this, you can write the series as $$\sum_{n=M+1}^\infty (-1)^n\frac{\sqrt n}{n+907}=\sum_{k=1}^M(-1)^k\frac{\sqrt k}{k+907}+\sum_{n=M+1}^\infty (-1)^n\frac{\sqrt n}{n+907}$$
and prove convergence by applying Leibniz to the second sum.
Personally, the way I like to prove that $\frac{\sqrt{n}}{n+907}$ is eventually decreasing is by evaluating the sign of the following derivative $$\frac{d}{dx}\left[\frac{\sqrt x}{x+907}\right]=\frac{-\frac12\sqrt x+\frac{907}{\sqrt x}}{(x+907)^2}$$
which is eventually $<0$.
