Question

a sequence of non negative real numbers is given {An} n= 1-> infinity ..... such that the summation of the series converges . Prove that the summation of {An^2} also converges. would this be different if the original series contained negative numbers

Answer 1

if it is
\[\sum(a_n)^2\] then it converges by the comparison test, since \((a_n)^2<a_n\)

Answer 2

for n big enough

Answer 3

right

Answer 4

if it has negative number i would say no
for example
\[\sum\frac{(-1)^n}{\sqrt{n}}\] converges as it is alternating and the terms go to zero
but the squares are \(\frac{1}{n}\)so the sum would diverge

Answer 5

thank you :) i think comparison test was the key i was looking for. I had already figured that the squared summation had to be smaller than the original since i could only figure it working if we were squaring something like fractions in which they kept getting smaller, but didn't know how to "prove it".

Answer 6

yeah i think "comparison" is the way to go since \(a_n\to 0\) means after some \(n\) (\(a_n)^2<a_n\)

Answer 7

yw