Determine if this series $\sum_{n=1}^\infty \frac{u_n}{\sqrt{n}}$ (with $u_n$ periodic) is convergent/divergent Let $u_n$ be defined as follows: 
\begin{align}
u_{4k} &= 1 \\ 
u_{4k+1} &= -2 \\ 
u_{4k+2} &= 2 \\ 
u_{4k+3} &= -1
\end{align}
Is the series $\sum_{n=1}^\infty \frac{u_n}{\sqrt{n}}$ convergent or divergent? 
I think it is convergent because you can compare it to a p-series with p = 3/2? But I am having trouble formalising the proof - I also tried to use something like the ratio test but I am dubious about whether this is a valid approach
 A: If you not pay attention to the very first term of the series (irrelevant for convergence) you can group your terms as 
$$
u_{4k+1}+u_{4k+2}+u_{4k+3}+u_{4k+4}.
$$
That is, 
\begin{align}
-\frac2{\sqrt{4k+1}}+\frac2{\sqrt{4k+2}}-\frac1{\sqrt{4k+3}}+\frac1{\sqrt{4k+4}}.
\end{align}
Putting the first two and the last two together, and rationalizing, you get 
$$
\frac{-2}{\sqrt{4k+1}\sqrt{4k+2}(\sqrt{4k+1}+\sqrt{4k+2})}+\frac1{\sqrt{4k+3}\sqrt{4k+4}(\sqrt{4k+3}+\sqrt{4k+4})}.
$$
This is  bounded above (via using $4k+1>4k$, $4k+2>4k$, etc. and looking at the second term) by 
$$
\frac1{16k^{3/2}}
$$
and below (looking at the first term) by 
$$
-\frac1{8k^{3/2}}
$$
So the series is convergent by comparison. 
A: Consider the series defined by
\begin{align}
v_{4k} &= 1 \\ 
v_{4k+1} &= 0\\ 
v_{4k+2} &= 0 \\ 
v_{4k+3} &= -1
\end{align}
and \begin{align}
w_{4k} &= 0 \\ 
w_{4k+1} &= -2 \\ 
w_{4k+2} &= 2 \\ 
w_{4k+3} &= 0
\end{align}
Then both $\sum_{n=1}^\infty \frac{v_n}{\sqrt{n}}$ and $\sum_{n=1}^\infty \frac{w_n}{\sqrt{n}}$ converge by an alternating series argument, and your series is their sum.
A: You want to group 4 consecutive terms together, which is doable since the terms tend to zero:
$$\sum_{k=0}^\infty-\frac2{\sqrt{4k+1}}+\frac2{\sqrt{4k+2}}-\frac1{\sqrt{4k+3}}+\frac1{\sqrt{4k+4}}$$
Note then that we have by the mean value theorem:
$$\frac1{(4k+2)^{3/2}}\le\frac2{\sqrt{4k+1}}-\frac2{\sqrt{4k+2}}\le\frac1{(4k+1)^{3/2}}$$
$$-\frac{1/2}{(4k+3)^{3/2}}\le-\frac1{\sqrt{4k+3}}+\frac1{\sqrt{4k+4}}\le-\frac{1/2}{(4k+4)^{3/2}}$$
And hence it converges by comparing to $\sum\frac1{k^{3/2}}$.
A: Note that we can write
$$\begin{align}
\sum_{n=1}^{4N}\frac{u_n}{\sqrt n}&=\sum_{n=1}^{N} \left(\frac{u_{4n-3}}{\sqrt{4n-3}}+\frac{u_{4n-2}}{\sqrt{4n-2}}+\frac{u_{4n-1}}{\sqrt{4n-1}}+\frac{u_{4n}}{\sqrt{4n}}\right)\\\\
&=\sum_{n=1}^{N} \left(\frac{-2}{\sqrt{4n-3}}+\frac{2}{\sqrt{4n-2}}+\frac{-1}{\sqrt{4n-1}}+\frac{1}{\sqrt{4n}}\right)\\\\
&=2\sum_{n=1}^N \left(\frac{1}{\sqrt{4n-2}}-\frac{1}{\sqrt{4n-3}}\right)+\sum_{n=1}^N \left(\frac{1}{\sqrt{4n}}-\frac{1}{\sqrt{4n-3}}\right)\tag1
\end{align}$$
The summands of both of the sums on the right-hand of $(1)$ are of order $O\left(n^{-3/2}\right)$.  Hence, the original series of interest converges.  And we are done!
