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I'm having trouble with a practice problem in sequences and series and would like some guidance. The problem is:

Given a series of real numbers $\sum_{k=1}^\infty x_k$, let $\sum_{n=1}^\infty y_n$ be another series of real numbers such that :

$y_1 = x_1$

$y_2 = \frac{x_2}{2}$


$\displaystyle y_n = \frac{1}{n(n-1)}x_2 + \frac{2}{n(n-1)}x_3 + \ldots \frac{n-2}{n(n-1)}x_{n-1} + \frac{1}{x_n}$

or, more concisely:

$\displaystyle y_n = \sum_{k=2}^n\frac{(k-1)}{n(n-1)}x_k$

The problem asks to prove if $\sum_{k=1}^\infty x_k$ converges, then $\sum_{n=1}^\infty y_n$ also converges $\textit{to the same value}$.

I can easily use the comparison test to prove that if one series converges, then the other must converge as well. However, I have no idea how to go about showing that both series converge to the same value.

One unsuccessful approach that I tried was to show that the partial sum $s_n = \sum_{k=1}^n$ and $t_n = \sum_{k=1}^n y_k$ were equivalent for all $n$. In the process of attempting this, I think that I actually managed to show that they were not....

This makes sense because my Prof. also gave us a hint saying that the second series may converge even though the first may not, and that the result can be used to say something along the lines of:

$\displaystyle 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 \ldots = \frac{1}{2}$

which makes absolutely no sense at all.

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up vote 1 down vote accepted

Let's consider a partial sum of the $y_n$: $$\sum_{n=1}^Ny_n = x_1 + \sum_{n=1}^N\sum_{k=2}^n\frac{k-1}{n(n-1)}x_k.$$ We'll use two basic ideas: first, that $\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$, so there's telescoping in this sum, and second, that we can interchange the order of summation, since $$\{(n,k): 1\le n\leq N, 2\leq k\leq n\ = \{(n,k): 2\leq k\leq N, k\leq n\leq N\}.$$

Applying both of these ideas, we get $$\begin{align*}\sum_{n=1}^Ny_n &= x_1 + \sum_{k=2}^N\sum_{n=k}^N\left(\frac{1}{n-1}-\frac1n\right)(k-1)x_k\\ &=x_1 + \sum_{k=2}^N(k-1)x_k\sum_{n=k}^N\left(\frac{1}{n-1}-\frac1n\right)\\ &=x_1 + \sum_{k=2}^N(k-1)x_k\left(\frac{1}{k-1}-\frac1N\right)\\ &= \sum_{k=1}^Nx_k - \sum_{k=2}^N\frac{k-1}{N}x_k. \end{align*}$$ So the difference between $\sum_{k\leq N}x_k$ and $\sum_{n\leq N}y_n$ is $\sum_{ k\leq N}\frac{k-1}{N}x_k$; we must show that the latter converges to $0$ as $N\to\infty$. If $\sum_n x_n$ is absolutely convergent, this is not too hard: the idea here is that the initial terms of the sum are small because $(k-1)/N$ is small, and the final terms of the sum are small because $x_k$ is small. For any $\epsilon>0$, choose $N_0$ such that $\sum_{k=n}^\infty |x_k|<\epsilon$ for all $n\geq N_0$. Then $\sum_{k=N_0}^{N}|x_k|<\epsilon$ for any $N\geq N_0$, so $\sum_{k=N_0}^{N}\frac{k-1}{N}|x_k|<\epsilon$ since each $(k-1)/N$ is less than $1$. Denote $|\sum_{k<N_0}(k-1)x_k|$ by $C$, which depends only on $\epsilon$. Then if we choose $N$ to be larger than $\max\{N_0, C/\epsilon\}$, we have $|\sum_{k<N_0}\frac{k-1}Nx_k| = C/N < \epsilon$. So $$\left|\sum_{k<N}\frac{k-1}{N} x_k\right| \leq \left|\sum_{k<N_0}\frac{k-1}{N} x_k\right| + \left|\sum_{k=N_0}^N\frac{k-1}{N} x_k\right|<\epsilon+\epsilon.$$ So for any $\epsilon$, the sum is eventually smaller than $2\epsilon$, so it converges to $0$.

I don't see immediately how to patch that last bit of the argument in case $\sum_nx_n$ is conditionally convergent - can anyone see how to proceed?

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