Test the following series for convergence or divergence. Justify the decision.

Let

Then, the series converges by Example #1 on page 398 of Apostol,

where (since ). Then consider the limit,

The limits of each of the terms in the product exist (as we show below) so the limit of the product is the product of the limits,

The limit since we know

for all , . Therefore, we have

By Theorem 10.9 (see the note after the proof of the theorem on page 396 of Apostol), we then have the convergence of implies the convergence of . Hence,

converges.

Well as n -> infinity

Sqrt(2n-1) ~ sqrt(2n)

n(n+1) ~ n^2

log(4n+1) ~ log(4n) = log(4) + log(n)

So just solve for

ln4 * Sqrt(2n)/n^2 + sqrt(2)*ln(n)/n^(3/2)

Which clearly converges for both parts -> 0

So the sum must converge by theorem 10.10… right?

Funny how Apostol clearly warned the reader from falling in this fallacy, if the terms of a series tend to zero, that doesn’t mean the series converges. The harmonic series is the classic example.

This can be done also simpler (from my point of view): notice that the logarithm term is less than n^{1/8} for all n > some N. Then, sqrt{2n – 1} < sqrt{2n}. From this, we find the identity a_n < \sqrt{2}b_n, which leads to Riemann zeta with power 11/8. Thus, by comparison test the series is convergent.