Series (20 pts)

a) Suppose that convergence of {(∞Σ n=1)an} can be established by ratio test. Prove that {(∞Σn=1)√(an)} converges. (10 marks)

As convergence of {(∞Σ n=1)an} can be established by ratio test,
Let L = lim (n--> inf) | an+1/an| >1
Then
lim(n-->inf) |√(an+1)/√(an)|
= √{lim (n--> inf) | an+1/an|}
= √L > 1
Hence {(∞Σn=1)√(an)} converges

b) If the convergence of {(∞Σ n=1)an} can be established by ratio test. Is it true that {(∞Σn=1)n^√(an)} will always converge？Give reason to justify your assertion. (10 marks)
remarks : n^√(an) = (an)^1/n
As convergence of {(∞Σ n=1)an} can be established by ratio test,
Let L = lim (n--> inf) | an+1/an| <1
Then
lim(n-->inf) |n^√(an+1)/n^√(an)|
= {lim (n--> inf) | n^√(an+1/an)|
= 1 when n tends to infinity
Hence {(∞Σn=1)n^√(an)} will not always converge

Part (a) is simple.
If the convergence of the series can be established by the ratio test, limit of |a_(n+1)/a_n| must be strictly smaller than 1. Since square root is a continuous function over (0, ∞), therefore, the
limit of √a_(n+1)/√a_n
= limit of √[a_(n+1)/a_n ]
= √limit of [a_(n+1)/a_n ]
&lt; 1.

Thus, by ratio test, the convergency of the required series converge.

For the part (b), it would be very similar except that we cannot claim the limit so easily because of the n itself make the function. For this part, the friend above answered that it may not always converge, can it be an &quot;assertion&quot; enough strength for this question? Also, it is not immediately clear that the last limit can be claimed. I believe for this question, we need to find a sequence that disprove the claim, for now I don&#39;t have such a sequence now.