Need help to compute the limit of a series

Question

So I have the following exercise:

exercise photo

I have determined that the series is convergent. But I do not know how to compute the limit.

Answer 1

$\sum_{n=0}^\infty{x^n}$ =$\frac{1}{1-x}$ . valid for -1< x<1 . One can differentiate the series term by term at x =$\frac{1}{2}$ that the sum of this series that you are asking about is $\frac{1}{(1-x)^2}$ which at x=$\frac{1}{2}$ is 4

Answer 2

The series is convergent by ratio test. For any convergent series $\sum a_n$ of positive terms $\lim \sum_n ^{2n} a_k =0$. In fact $\lim \sum_n ^{\infty} a_k =0$.

Answer 3

Clearly for $k=n,\dotsc, 2n$, $n\leq k \leq 2n$, so $$\sum_{k=n}^{2n}\frac{n}{2^{k-1}} \leq \sum_{k=n}^{2n}\frac{k}{2^{k-1}} \leq \sum_{k=n}^{2n}\frac{2n}{2^{k-1}}$$ and also note that $$\sum_{k=n}^{2n}\frac{1}{2^{k-1}} = \frac{1}{2^{n-1}}\sum_{k=0}^{n}\frac{1}{2^k} = \frac{1}{2^{n-1}} \frac{1-(1/ 2)^{n+1}}{1- (1/ 2)} = \frac{4}{2^n} (1- (1/ 2)^{n+1})$$ then $$\underbrace{\frac{4 n}{2^n}}_{\to 0} \cdot \underbrace{(1-(1/ 2)^{n+1})}_{\to 1} = \sum_{k=n}^{2n}\frac{n}{2^{k-1}} \leq \sum_{k=n}^{2n}\frac{k}{2^{k-1}} \leq \sum_{k=n}^{2n}\frac{2n}{2^{k-1}} = \underbrace{\frac{8 n}{2^n}}_{\to 0} \underbrace{(1-(1/ 2)^{n+1})}_{\to 1}$$

So the limit is zero by the squeeze theorem.