# Thread: Does this series converge?

1. ## Does this series converge?

Let $g_k(x)= \left\{\begin{array}{cc} \frac {1}{k^2},&\mbox{ if }
|x| \leq k\\ \frac {1}{x^2}, & \mbox{ if } |x|>k\end{array}\right.$

Does the series $\sum ^ \infty _{k=1} g_k(x)$ converges pointwise or uniformly?

Proof so far.

Define the partial sum $s_n= \sum ^n_{k=1} g_k (x)$.

Now, in the case that $|x| \leq k$, we have $s_n(x)= \sum ^n_{k=1} \frac {1}{k^2} = \sum ^n_{k=1} ( \frac {1}{k})^2$

Now, does this one converges to $\frac {1}{k^2}$, if it does then it is pointwise.

In the case that $|x| > k$, then $s_n = \sum ^ n_{k=1} \frac {1}{x^2} = \frac {n-1}{x^2} = \frac {n}{x^2} - \frac {1}{x^2}$

Well, then this guy doesn't converge to $g_k$, so it ain't pointwise convergence then?

Thanks, people, I'm really lost in series convergence here...

Let $g_k(x)= \left\{\begin{array}{cc} \frac {1}{k^2},&\mbox{ if }
Does the series $\sum ^ \infty _{k=1} g_k(x)$ converges pointwise or uniformly?
if $|x| > k,$ then $g_k(x)=\frac{1}{x^2} < \frac{1}{k^2}.$ thus: $g_k(x) \leq \frac{1}{k^2},$ for all $x \in \mathbb{R}, \ k \in \mathbb{N}.$ hence by Weierstrass M-test your series is uniformly convergent. Weierstrass M-test