sequence/series limits I am struggling a bit with the following two questions I got from a sister who is doing BSC in Applied Mathematics. I always try my best to help her but was stuck on these questions and so any answer and explanation is much appreciated:
Here's the question:
Decide which of the following sequences have limits when $n\rightarrow \infty$ , and find the relevant limits if they exist. In the case of $C(n)$ it is enough to show that the limit is bounded from above and below.
Sequence $B(n)$ such that
$B(0) = 0$ and $B(n) = B(n-1) + 1/2^n$ , for any positive integer $n$.
Sequence $C(n)$ such that $C(1) = 1$ and $C(n) = C(n-1) + 1/n^2$ , for any integer $n>1$

(Hint: Use the fact $\frac{1}{n-1} -\frac{1}{n} =\frac{1}{(n(n-1))}$ with the sandwich reasoning that if $C'(n)<C(n)<C''(n)$ for all $n$, then the limit of $C(n)$ is between the limits of $C'(n)$ and $C''(n)$, if such limits exists.)

On the first question , I tried to plug some values of n and got the limit of B(n) as 1, although I am not 100% sure. But on the second one, I couldn't do much and the hint didn't help me a lot.

$B(0)=0$, $B(n)=B(n-1)+\frac 1 {2^{n}}$,
Thus $$B(1)= \frac 1 2,B(2)= \frac 1 2+\frac 1 4,B(3)=\frac 1 2+ \frac 1 4+\frac 1 8,\cdots B(n)= \frac1 2+\cdots \frac 1 {2^{n}} $$
Now see $$\frac{1}{2} B(n)= \frac 1 4+\frac1 8+\cdots \frac 1 {2^{n+1}}\\\implies \left(1-\frac 1 2\right)B(n)=\frac 1 2-\frac 1 {2^{n+1}}\\\implies \lim_{n\to +\infty}B(n)=1$$
$C(1)=1,C(n)=C(n-1)+\frac 1 {n^2}$
Similarly we get, $C(n)=1+\frac 1 {2^2}+\frac 1 {3^2}+\cdots +\frac 1 {n^2}$
Observe $$\frac 1 {2^2}+\frac 1 {3^2}<\frac 2 {2^2}=\frac 1 {2}\\\frac 1 {4^2}+\frac 1 {5^2}+\frac 1 {6^2}+\frac 1 {7^2}<\frac 4 {4^2}=\frac 1 {4}\\\cdots\\$$
Similarly we can say that $C_{2^{n+1}-1}<1+\frac 1 {2}+\frac 1 {4}+\cdots +\frac 1 {2^n}=B(n)+1$
Therefore $\lim C(n) <\lim(B(n))+1$, Using the fact that $C(n)$ is increasing you can conclude that $\lim C(n)$ exists.
