In some example tasks for infinite function series I'm learning from pointwise convergence is proven like we are dealing with sequence instead of series - $\lim_{n \rightarrow \infty}f_n(x)$ when , result being limiting function of a given series.
So, for example if following series is given:
$$ \sum _{n=1}^{\infty} \frac{x}{e^{n^{2}x}} $$
Pointwise convergence is proven by:
$$\lim_{n \rightarrow \infty}\frac{x}{e^{n^{2}x}} = 0$$
I'm not sure if that is correct, anyone could clarify?
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For a series to converge, convergence of the sequence to zero is an obvious necessary condition. But it is not sufficient: if the terms decrease slowly, the sum can be unbounded.
Consider the sequence
$$1,\frac12,\frac12,\frac14,\frac14,\frac14,\frac14,\frac18,\frac18,\frac18,\frac18,\frac18,\frac18,\frac18,\frac18,\cdots$$
Obviously it tends to zero, but the partial sums diverge.
To study convergence of a series, squeezing remains the basic technique and comparison to a geometric series leads to the ratio test.
Another important result is that
$$\sum_{n=1}^\infty\frac1{n^\alpha}$$ converges iff $\alpha>1$.
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For a sequence of functions $(f_k)_{k\in\mathbb N}$, pointwise convergence means that for every fixed $x$ the sequence of values $(f_k(x))_{k\in\mathbb N}$ converges.
In your case $$ f_k(x) = \sum_{n=0}^k \frac{x}{e^{n^{2}x}} $$ and pointwise convergence means that for all fixed $x$ the series $\sum_{n=0}^\infty \frac{x}{e^{n^{2}x}}$ converges.
Convergence of the n-th term of $\sum a_n$ to $0$ does not guarantee convergence of the series. The given series is not convergent if $x<0$. For $x=0$ it is obviously convergent. For $x>0$ apply ratio test to prove that it is convergent. [ $\lim \frac {a_{n+1}} {a_n}=\lim e^{-2nx} e^{-x} =0 <1$ so the series is convergent.