$ 1+(1/\sqrt{3})-(a/\sqrt{2})+(1/\sqrt{5})+(1/\sqrt{7})-(a/\sqrt{4})+(1/\sqrt{9})+(1/\sqrt{11})-(a/\sqrt{6})... $ Converges By comparison test?

Question

For what value of real constant $a$ does the following series converge? $$ 1+(1/\sqrt{3})-(a/\sqrt{2})+(1/\sqrt{5})+(1/\sqrt{7})-(a/\sqrt{4})+(1/\sqrt{9})+(1/\sqrt{11})-(a/\sqrt{6})... $$ I do not have a clue on how to proceed. Suggest a possible route.

can we do it by comparison test somehow? Can we comapre it with $c/\sqrt{(n)}$ for some $c$ and some large enough $n$?

Answer 1

So, the terms come in blocks of three, where the sum of the $k$-th block of terms is $$ s_k=y_{3k-2}+y_{3k-1}+y_{3k}=\frac{1}{\sqrt{4k-3}}+\frac{1}{\sqrt{4k-1}}-\frac{a}{\sqrt{2k}}. $$ Since the individual $y_i\rightarrow 0$, the sum $\sum_{k=1}^{\infty} y_k$ converges if and only if the sum $\sum_{k=1}^{\infty} s_k$ converges. I suggest you calculate the asymptotic behavior of $s_k$. For large $k$, $$s_k \sim A(a) k^{-1/2} + B(a) k^{-3/2}+O(k^{-5/2}),$$ so the sum converges if and only if $A(a)=0$. Can you find what this implies about $a$?

Answer 2

Okay, I think I can guess at the pattern with the most recent addition $$ \begin{align} &\sum_{n=0}^\infty\left(\frac1{\sqrt{4n+1}}+\frac1{\sqrt{4n+3}}-\frac{a}{\sqrt{2n+2}}\right)\\ &=\sum_{n=0}^\infty\frac{\sqrt{(4n+3)(2n+2)}+\sqrt{(4n+1)(2n+2)}-a\sqrt{(4n+1)(4n+3)}}{\sqrt{(4n+1)(4n+3)(2n+2)}}\\ &=\sum_{n=0}^\infty\frac{\sqrt{(1+\frac3{4n})(1+\frac1n)}+\sqrt{(1+\frac1{4n})(1+\frac1n)}-a\sqrt2\sqrt{(1+\frac1{4n})(1+\frac3{4n})}}{2\sqrt{n}\sqrt{(1+\frac1{4n})(1+\frac3{4n})(1+\frac1n)}}\\ &=\sum_{n=0}^\infty\frac{\left[1+O(\frac1n)\right]+\left[1+O(\frac1n)\right]-a\sqrt2\left[1+O(\frac1n)\right]}{2\sqrt{n}\left[1+O(\frac1n)\right]}\\ &=\sum_{n=0}^\infty\frac{2-a\sqrt2+O(\frac1n)}{2\sqrt{n}}\tag{1} \end{align} $$ where $O(\cdot)$ is Landau big-O notation. Since convergence concerns the tail of the sequence, we can replace the term for $n=0$ by it value, $1+\frac1{\sqrt3}-\frac{a}{\sqrt2}$, and start using $O(\frac1n)$ for $n\ge1$.

The sum in $(1)$ will converge if and only if $a=\sqrt2$.

Answer 3

Expanding on mjqxxxx's answer:

$s_k=y_{3k-2}+y_{3k-1}+y_{3k}=\frac{1}{\sqrt{4k-3}}+\frac{1}{\sqrt{4k-1}}-\frac{a}{\sqrt{2k}}.$

By the binomial theorem,

$\begin{array}\\ (4k-c)^{-1/2} &=(4k)^{-1/2}(1-\frac{c}{4k})^{-1/2}\\ &=(4k)^{-1/2}(1-\frac{c}{4k}\binom{-1/2}{1}+O(k^{-2}))\\ &=(4k)^{-1/2}(1+\frac{c}{8k}+O(k^{-2}))\\ \end{array} $

Therefore

$\begin{array}\\ (4k-1)^{-1/2}+(4k-3)^{-1/2} &=(4k)^{-1/2}(2+\frac{4}{8k}+O(k^{-2}))\\ &=k^{-1/2}+\frac{1}{4k^{3/2}}+O(k^{-2}))\\ \end{array} $

To get the $k^{-1/2}$ term to get cancelled, we must have $\frac{a}{\sqrt{2}} =1 $ or $a = \sqrt{2}$.