Question

Given that \[\sum_{n=1}^{\infty}u_n\] is a convergent series of positive terms, show that \[\sum_{n=1}^{\infty}u_n^p\] converges for \[p>1\].

Here is my attempt:
Let \[p>1\]. Then if \[0 14 years ago

Answer 1

\[\Longrightarrow \sum_{n=1}^{\infty}\] converges.

Answer 2

\[\Longrightarrow \sum_{n=1}^{\infty}u_n^p\] converges *

Answer 3

This is my attempt, I wonder if it's right or wrong?

Answer 4

No... well, convergence would be, stated formally, with Sn as your sum
\[\forall \epsilon \in \mathbb{R}^{+}, \exists m \in \mathbb{N}, n>m \Rightarrow |Sn-l|<\epsilon \]
Now if your first series converges, it means the more terms you add, the more small or "epsilonesque" these terms become, because the previous definition of convergence would be verified. Now, if those really small terms that are eventually in the neighborhood of 0 are elevated to the power of p>1, those terms will get smaller, that's why it converges... Because if it converges
\[\exists m \in N, n>m \Rightarrow |Sn-l| \in V(0)\]
where V(0) is a neighborhood of 0 defined as
\[\nu \in \mathbb{R}^{+}, V(0)=]-\nu;\nu[\]
for any arbitrarily small nu.
Now as your counter gets larger, your terms are in a sufficiently small neighborhood of 0 so that when you elevate them to the p they're in an even "smaller" (in terms of inclusion) neighborhood.. i.e. the terms you add become smaller so it converges...

Answer 5

@someone, for all that formalism, you still haven't said anything yet about Callum's proof.

Answer 6

You're adding the terms of a series, the individual terms may converge but that doesn't mean the sum itself does

Answer 7

i think maybe it would be easier to start by asserting that since
\[\sum u_n\] converges, it is necessary that
\[\lim_{n\rightarrow \infty}u_n=0\] and so there exists and N such that for all n > N
\[u_n<1\]
then you can skip the the second argument altogether

Answer 8

* an N

Answer 9

in fact i am not sure the second part of what you wrote makes sense. you are given that
\[\sum u_n\] is finite, and you want to show that
\[\sum u_n^p\] is finite as well. comparison test does it for you. it cannot be the case that
\[u_i>1\] for more than a finite number of terms

Answer 10

Thanks Satellite, that's really helped. I had not thought about \[\sum_{n=1}^{\infty}u_n\] converges \[\Longrightarrow \lim_{n\rightarrow \infty}u_n = 0\]

Answer 11

yw, hope it is clear

Answer 12

If one wanted to prove that fact I guess it goes like this?\[\sum_{n=1}^{\infty}u_n\] converges \[\Longrightarrow \lim_{n\rightarrow \infty}s_n = l\] exists, where \[s_n\] is the nth partial sum.
For N sufficiently large, we have that \[|s_n - l| < \varepsilon\] \[\forall n>N\].
So \[s_{n+1} - s_n = u_{n+1} \Longrightarrow 0 = l-l = \lim_{n\rightarrow \infty}(s_{n+1}-s_n) = \lim_{n\rightarrow \infty}u_{n+1} \Longrightarrow \{u_{n}\}\] converges to 0.