Check convergence /divergence of a series $\sum_{n=1}^{\infty} (-1)^{n+1}(n)^{\frac{1}{4}}(\frac{1}{\sqrt{4n-1}}- \frac{1}{\sqrt{4n}} )$

Question

for $(a)$ and $(b)$ decide if the series converges , converges conditionally or diverge

$(a)$ $\sum_{n=1}^{\infty} (-1)^{n+1}(n)^{\frac{1}{4}}(\frac{1}{\sqrt{4n-1}}- \frac{1}{\sqrt{4n}} )$

$(b)\ $ $\sum_{n=1}^{\infty} (-1)^{n+1}(n)^{\frac{5}{4}}(\frac{1}{\sqrt{4n-1}}- \frac{1}{\sqrt{4n}} )$

i tried to use Leibniz test for $(a)$ and $(b)$ but failed. i tried to do assymptotic tests but failed.

also i am not sure how to decide if the sequence monotonically decreasing

Answer 1

Answer to first series: $\frac 1 {\sqrt {(4n-1)}}- \frac 1 {\sqrt {4n}} =\frac {\sqrt {4n}-\sqrt {4n-1}} {\sqrt {4n-1} \sqrt {4n}}$. Write this as $\frac 1 {\sqrt {4n-1} \sqrt {4n}(\sqrt {4n}+\sqrt {4n-1})}$. Note that the denominator is of the order of $n^{3/2}$. Apply comparison test now (Note that $\frac 3 2-\frac 1 4=\frac 5 4 >1)$. Hence the first series is absolutely convergent. A similar argument shows that the second series is not absolutely convergent. Try to prove its convergence using Alternating Series Test.