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# If each term in a sum converges, does the infinite sum converge too?

Let $S(x) = \sum_{n=1}^\infty s_n(x)$ where the real valued terms satisfy $s_n(x) \to s_n$ as $x \to \infty$ for each $n$.

Suppose that $S=\sum_{n=1}^\infty s_n< \infty$.

Does it follow that $S(x) \to S$ as $x \to \infty$?

I really do not recall any theorems... please point out !!!

This is not true. Consider $s_n(x) = 1/x$. We have $\lim_{x \to \infty} s_n(x) = 0 = t_n$. Clearly, $\sum_{n=1}^{\infty} t_n =0 < \infty$. However, $$S_n(x) = \sum_{k=1}^n s_n(x) = \dfrac{n}x$$ and $\lim_{n \to \infty} S_n(x)$ doesn't exist.

No.

Let $s_n(x)=\frac 1x$, $s_n=0$. Then $S=\sum_{n=1}^\infty s_n$ trivially converges. However, $S(x)=\sum_{n=1}^\infty s_n(x)$ is clearly divergent, since the summand is a constant (with respect to $n$) greater than $0$.

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# If each term in a sum converges, does the infinite sum converge too?

Let $S(x) = \sum_{n=1}^\infty s_n(x)$ where the real valued terms satisfy $s_n(x) \to s_n$ as $x \to \infty$ for each $n$.

Suppose that $S=\sum_{n=1}^\infty s_n< \infty$.

Does it follow that $S(x) \to S$ as $x \to \infty$?

I really do not recall any theorems... please point out !!!

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This is not true. Consider $s_n(x) = 1/x$. We have $\lim_{x \to \infty} s_n(x) = 0 = t_n$. Clearly, $\sum_{n=1}^{\infty} t_n =0 < \infty$. However, $$S_n(x) = \sum_{k=1}^n s_n(x) = \dfrac{n}x$$ and $\lim_{n \to \infty} S_n(x)$ doesn't exist.

So in your example $S(x)$ is infinite. If it is assumed finite, for every $x$, is it true? - RealMax yesterday

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