For every k &#x2208;<!-- ∈ --> <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-str

hornejada1c

hornejada1c

Answered question

2022-07-11

For every k N , let
x k = n = 1 1 n 2 ( 1 1 2 n + 1 4 n 2 ) 2 k .

Answer & Explanation

escampetaq5

escampetaq5

Beginner2022-07-12Added 12 answers

Solution 1: By the dominated convergence theorem, we may switch the order of the limit and the sum, so we see that
lim k x k = 0.
Solution 2: Notice the terms are always bounded above by 1 n 2 . Let ϵ > 0, and choose N such that
n = N 1 n 2 < ϵ .
Then choose k so large that
( 1 1 2 N ) 2 k ϵ N .
It then follows that | x k | 2 ϵ , and the same inequality holds for all j k. Since ϵ was arbitrary the proof is finished.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?