# 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

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.

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.

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 $$\sum\frac1{k^{3/2}}$$.

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!