Am I correct if I say $\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{\sqrt{k}}{(k+1)^2}$ converges? The question:
Test the following series for convergence or divergence:
$$
\frac{1}{2^2}+\frac{\sqrt{2}}{3^2}+\frac{\sqrt{3}}{4^2}+\frac{\sqrt{4}}{5^2}+\dots
$$
My solution
The general function would be $$\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{\sqrt{k}}{(k+1)^2}$$
so
$$\lim_{n\rightarrow\infty}\sqrt{n}=\infty$$
also
$$\lim_{n\rightarrow\infty}(n+1)^2=\infty$$
so the series is convergent because $\frac{\infty}{\infty}=1$, no?
 A: No. 
First, "$\infty/\infty=1$'' makes no sense. 
Second, even if it did, one could apply your argument to show that $\sum_{k=1}^\infty \frac{n}n$ converges.
The series you are considering converges because it is comparable with a convergent series. Namely,
$$
\frac{\sqrt n}{(n+1)^2}\leq\frac{\sqrt n}{n^2}=\frac1{n^{3/2}}.
$$
As $\sum_{k=1}^\infty n^{-3/2}<\infty$, we conclude by comparison that $\sum_{k=1}^\infty \sqrt{n}/(n+1)^2<\infty$.
A: Simply saying $\frac\infty\infty$ is meaningless. Does $\sum_{n=1}^\infty\frac{n}{n}$ converge? No. In this case, $$\lim_{n\to\infty}\frac{\sqrt{n}}{(n+1)^2}=0,$$ which is very important, since if these terms don't tend to $0$, then the series can't possibly converge.
In this case, you'll want to use direct comparison test and integral test, instead. $$0<\frac{\sqrt{n}}{(n+1)^2}<\frac{\sqrt{n+1}}{(n+1)^2}=\frac1{(n+1)^{3/2}},$$ So since $$\int_1^\infty\frac1{(x+1)^{3/2}}\,dx$$ converges (check), then $$\sum_{n=1}^\infty\frac1{(n+1)^{3/2}}$$ converges by the integral test, and so $$\sum_{n=1}^\infty\frac{\sqrt{n}}{(n+1)^2}$$ converges by direct comparison test.
A: The series indeed converges. Since $(n+1)^2 > n^2$, we have
$\sum_{n=1}^{\infty} \frac{\sqrt{n}}{(n+1)^2} < \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2} = \sum_{n=1}^{\infty} n^{\frac{-3}{2}}$.
Now it is known by integral test that $\sum_{n=1}^{\infty}\frac{1}{n^s}$ converges for $s>1 \in \mathbb{R}$.
A: $$0 \lt \lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{\sqrt{n}}{(n+1)^2} \lt \lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{\sqrt{n+1}}{(n+1)^2} \lt \lim_{n\rightarrow\infty}\int_{x=1}^{n+1} x^{-3/2} \, dx = 2. $$  So the series limit converges on  a number less than $2$ from below.
A: You can use Limit comparison test. Since $$\lim_{n \to \infty}\dfrac{\frac{\sqrt{n}}{(n+1)^2}}{\frac{1}{n^\frac{3}{2}}}=1$$
and the series $\displaystyle{\sum_{k=1}^\infty\frac{1}{n^\frac{3}{2}}}$ converges, the series $\displaystyle{\sum_{k=1}^\infty\frac{\sqrt{n}}{(n+1)^2}}$ converges.
A: In short: yes, but not for that reason. What you are doing is not a valid test for convergence. For example, $a_n = \frac{n^2}{n}$ both top and bottom converge to $\infty$ yet the series certainly diverges. Any argument involving "$\frac{\infty}{\infty} = 1$" is likely to be wrong. 
However, the series does converge for several reasons. I like $a_n < \frac{\sqrt{n+1}}{(n+1)^2} = (n+1)^{-\frac{3}{2}}$ which converges since $\frac{3}{2} >1$, a well known result (which can be proved by the integral test)
