A tricky question involving convergence of series

Question

I'm struggling to find a title for this. Hopefully it can be edited subsequently.

Suppose $\sum_k s_k(x)$ converges to $S(x)$

Now suppose each $s_k(x)$ has a sequence $t_k^N(x)$ approaching it (as $N \to \infty$)

Can we say that $\sum_k{t_k^N(x)} \to S(x)$ as $N \to \infty$?

I'm pretty sure the answer is "not necessarily" and that uniform convergence is necessary.

Is this correct?

Answer 1

A simple example where $\sum t_k^N(x)$ diverges and $\sum a_k(x)$ converges is $t_k^N(x) = x(1/k^2 + 1/N).$

As an example where both series converge and all limits exist, but you cannot interchange the limit and sum, let $t_k^N(x) = \frac1{N}e^{-k/N}$.

Note that

$$\sum_{k=0}^\infty \lim_{N \to \infty}\frac1{N}e^{-k/N} = 0,$$

but

$$\lim_{N \to \infty}\sum_{k=0}^\infty \frac1{N}e^{-k/N} = \lim_{N \to \infty}\frac{1/N }{1 - e^{-1/N}} = 1.$$

Uniform convergence is a sufficient condition that permits the interchange, but it is not necessary.

To see that it is not necessary, consider the following counterexample:

$$t_1^N = \frac{1}{N}\left(1 - \frac{1}{N} \right); \,\,\, \,t_k^N =\frac{k}{N}\left(1 - \frac{1}{N} \right)^k - \frac{k-1}{N}\left(1 - \frac{1}{N} \right)^{k-1} \,\, \text{for }\,\, k = 2, 3, \ldots$$

Then

$$a_k = \lim_{N \to \infty}t_k^N = 0, \\ S_n^N = \sum_{k=1}^n t_k^N = \frac{n}{N}\left(1 - \frac{1}{N} \right)^n, \\ \sum_{k=1}^\infty t_k^N = \lim_{n \to \infty}S_n^N = 0,$$

and

$$\lim_{N \to \infty} \sum_{k=1}^\infty t_k^N = 0 = \sum_{k=1}^\infty \lim_{N \to \infty} t_k^N.$$

However, $\sum t_k^N$ does not converge uniformly to $0$ for $N \in \mathbb{N}, $ since with $N = n$

$$\lim_{n \to \infty}S_n^n = \lim_{n \to \infty}\left(1 - \frac{1}{n} \right)^n = e^{-1} \neq 0.$$

Answer 2

Let $t_k^N=1$ for $N\leq k$ and let $t_k^N=0$ when $N>k. $ Then $s_k=\lim_{N\to \infty}t_k^N=0. $ So $\sum_{k\in N}s_k=0. $ But for $n>N$ we have $\sum_{j=1}^nt_j^N=n-N. $ So $\sum_{j\in N}t_j^N=\infty $ for each $N.$