# Thread: Convergence- proving bounded

1. ## Convergence- proving bounded

Given sequence, a= 1/1^2+1/2^2+1/3^2+.......+1/n^2

(1) I have to prove for natural numbers n, such that n is greater than or equal to 2,

a is less than or equal to 1+1/1*2+1/2*3+1/3*4+.....+1/(n-1)*n

(2) Prove that a is bounded for natural numbers

I've been given a hint in the question to use the fact that for n greater than or equal to 2, 1/(n-1)n=1/(n-1) -1/n

Anyone have any ideas? Anything would be very much appreciated

2. Hello.

In part one, you can just compare terms directly. The $n$-term in your first series is $\frac{1}{n^2}$, and the $n$-term in your second series is $\frac{1}{(n-1)n}$. Is it true that $\frac{1}{n^2}\leq\frac{1}{(n-1)n}$ (for $n>2$)? If so, then you have $1+\frac{1}{2^2}+\ldots+\frac{1}{n^2}<1+\frac{1}{1\ cdot 2}+\ldots+\frac{1}{(n-1)n}$, and therefore, the first series is bounded by the second.

In part two, we can try to sum up the second series to put a bound on the first. Try writing it down using your hint:

$1+\frac{1}{1\cdot 2}+\ldots+\frac{1}{(n-1)n}=1+\frac{1}{2}+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\ldots+(\frac{1}{n-2}-\frac{1}{n-1})+(\frac{1}{n-1}-\frac{1}{n})$

Anything nice happen? What happens when you take $n\to \infty$? Use this to conclude.