Series Multiplication: Investigate Convergence

[Lisa91]

I multiplied two series $\displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n^{\frac{3}{4}}}$ and $\displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n^{\frac{1}{4}}}$. I got $\displaystyle 1 + \frac{1}{2^{\frac{3}{4}}2^{\frac{1}{4}}}+\cdots$ I don't know how to write it with symbols so that I could investigate whether it's convergent or not. I'd be thankful for your help!

 

[Sherlock1]

I multiplied two series $\displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n^{\frac{3}{4}}}$ and $\displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n^{\frac{1}{4}}}$. I got $\displaystyle 1 + \frac{1}{2^{\frac{3}{4}}2^{\frac{1}{4}}}+\cdots$ I don't know how to write it with symbols so that I could investigate whether it's convergent or not. I'd be thankful for your help!

Welcome. :] I fixed your latex. I hope the following is what you were after. Using the product formula...$\begin{aligned} \bigg(\sum_{n=0}^{\infty} (-1)^n n^{-3/4}\bigg) \bigg(\sum_{n=0}^{\infty} (-1)^n n^{-1/4}\bigg) = \sum_{n=0}^{\infty}\sum_{k=0}^{n}(-1)^k k^{-3/4}(-1)^{n-k}(n-k)^{-1/4} = \sum_{n=0}^{\infty}\sum_{k=0}^{n}(-1)^n k^{-3/4}(n-k)^{-1/4}\end{aligned}$

 

[Lisa91]

First of all, may I write it like this $$\sum_{n=0}^{\infty}(-1)^{n}\sum_{k=0}^{n} k^{-3/4}(n-k)^{-1/4}$$?
$$\sum_{k=0}^{n} k^{-3/4}(n-k)^{-1/4}$$ I read that if I want to investigate whether the whole series is convergent or not I have to investigate this guy at first.
I think I could compare this guy with something that I know but I am a bit confused about 'n' and 'k'. Should I care about 'n' in this one. May I assume that it's fixed?

Then I'll investigate: $$\sum_{n=0}^{\infty}(-1)^{n}\sum_{k=0}^{n} k^{-3/4}(n-k)^{-1/4}$$. I could apply the Leibniz formula but once again what about 'n' and 'k'.

 

[Opalg]

First of all, may I write it like this $$\sum_{n=0}^{\infty}(-1)^{n}\sum_{k=0}^{n} k^{-3/4}(n-k)^{-1/4}$$?
$$\sum_{k=0}^{n} k^{-3/4}(n-k)^{-1/4}$$ I read that if I want to investigate whether the whole series is convergent or not I have to investigate this guy at first.
I think I could compare this guy with something that I know but I am a bit confused about 'n' and 'k'. Should I care about 'n' in this one. May I assume that it's fixed?

Then I'll investigate: $$\sum_{n=0}^{\infty}(-1)^{n}\sum_{k=0}^{n} k^{-3/4}(n-k)^{-1/4}$$. I could apply the Leibniz formula but once again what about 'n' and 'k'.

For $x$ in the interval $0\leqslant x\leqslant 1$, the function $f(x) = x^{3/4}(1-x)^{1/4}$ has a maximum value $3^{3/4}/4$ (which occurs when $x=3/4$). Putting $x=k/n$, you see that $k^{3/4}(n-k)^{1/4} \leqslant 3^{3/4}n/4$. Therefore $\displaystyle\sum_{k=0}^{n} k^{-3/4}(n-k)^{-1/4} \geqslant \frac{4(n+1)}{3^{3/4}n}>1$.

 

[blue_raver22]

Hello,

Thank you for sharing your findings with me. It appears that you have multiplied two series, $\displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n^{\frac{3}{4}}}$ and $\displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n^{\frac{1}{4}}}$, and obtained a new series, $\displaystyle 1 + \frac{1}{2^{\frac{3}{4}}2^{\frac{1}{4}}}+\cdots$. To investigate whether this new series is convergent or not, we can use the ratio test or the root test.

The ratio test states that if $\displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$, then the series $\displaystyle \sum_{n=1}^{\infty} a_n$ is convergent. Similarly, the root test states that if $\displaystyle \lim_{n \to \infty} \sqrt[n]{|a_n|} < 1$, then the series $\displaystyle \sum_{n=1}^{\infty} a_n$ is convergent.

In this case, we can apply the ratio test, since the terms of the series are all positive. We have $\displaystyle \frac{a_{n+1}}{a_n} = \frac{\frac{1}{(n+1)^{\frac{3}{4}}}}{\frac{1}{n^{\frac{3}{4}}}} = \left( \frac{n}{n+1} \right)^{\frac{3}{4}}$. Taking the limit as $n \to \infty$, we get $\displaystyle \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^{\frac{3}{4}} = \lim_{n \to \infty} \left( \frac{1}{1 + \frac{1}{n}} \right)^{\frac{3}{4}} = 1^{\frac{3}{4}} = 1$. Since this limit is equal to 1, the ratio test is inconclusive and we