For each of the statements below, prove the result if it is true, give a counter example if it is not true

Question

For each of the statements below, prove the result if it is true, give a counter example if it is not true.

Suppose $\sum a_n$ with $a_n> 0$ is convergent. Then,

a) $\sum \sqrt{a_na_{n+1}}$ is convergent

b) for all $0< \delta < 1$, $\sum a_n^{1+\delta }$ is convergent

c) for all $0< \delta < 1$, $\sum a_n^{1-\delta }$ is convergent

I tried taking out cases for each by having $a_n=1/n$ which converges but unable to proceed further. For part b, I think it should not converge always since the power gets more than 1, so there may be cases where it doesn't converge. But I'm unable to find any case. Can anyone help out ?

Thanks in advance!

Answer 1

For Part C consider the series summation 1/n^(1.2) then this series is convergent by P test , but if we choose delta to be - 0.3 then series summation 1/n^(1.2)*0.7 which is series summation 1/n^(0.84) which is divergent by P test . so , C is false , for part B since delta is in the range 0 to 1 so , 1+ delta is a number between 1 and 2 , also since series a_n converges so ,a_n tends to zero as n tends towards infinity that means there exists a Natural number N such that 0 =N (Taking epsilon to be 1/2 )so , (a_n)^(1+delta)< a_n for all n >= N so, summation (a_n)^(1+delta)< summation a_n where summation runs over N to infinity , since summation a_n converges so , summation (a_n) is finite for n>=N , so , for all n>= N summation (a_n)^(1+delta) is finite ,so , whole series converges. Part B is correct, for part A -use GM<= AM since sequence terms are non negative we can use this - so , square root (a_n * a_n+1)<= 1/2 *(a_n+a_n+1)< a_n+a_n+1 , so , summation (a_n * a_n+1)<= summation (a_n+a_n+1) = summation a_n + summation a_n+1 which is finite , so , series converges by comparison test .