Does the series converge, problem with $(-1)^n$, but Leibnitz impossible.

Question

Hi my problem is connected with this series $$ \sum_{n=2}^\infty \frac{(-1)^n}{\sqrt[3]n + (-1)^{n(n+1)/2}} $$ I was trying to sum it by $4$ elements, but it didn't end up well. I don't know how to deal with this problem, Leibnitz, Dirichlet, Abel is pointless here. Can you please help me?

Answer 1

We consider the series $$ \sum\limits_{n = 4}^{ + \infty } {\frac{{\left( { - 1} \right)^n }} {{\sqrt[3]{n} + \left( { - 1} \right)^{\frac{{n\left( {n + 1} \right)}} {2}} }}} $$ instead your series. If we prove that this series is convergent then your series is convergent also. We can rewrite the series as $$ \sum\limits_{k = 1}^\infty {\left( {a_{4k} + a_{4k + 1} + a_{4k + 2} + a_{4k + 3} } \right)} $$ where $$ a_{4k} = \frac{1} {{\sqrt[3]{{4k}} + 1}},a_{4k + 1} = - \frac{1} {{\sqrt[3]{{4k + 1}} - 1}},a_{4k + 2} = \frac{1} {{\sqrt[3]{{4k + 2}} - 1}},a_{4k + 3} = - \frac{1} {{\sqrt[3]{{4k + 3}} + 1}} $$ Now, write down $$ s_1 \left( k \right) = \frac{1} {{\sqrt[3]{{4k}} + 1}} - \frac{1} {{\sqrt[3]{{4k + 1}} - 1}} $$ and rewrite is as function of $\frac{1}{k}$. After that using the Taylor series also, you can develop this function in powers of $\frac{1}{k}$. You will get $$ s_1 \left( k \right) = - \frac{{\left( {\frac{1} {k}} \right)^{2/3} }} {{\sqrt[3]{2}}} - \frac{{5\left( {\frac{1} {k}} \right)^{4/3} }} {{12\;2^{2/3} }} + \frac{{\left( {\frac{1} {k}} \right)^{5/3} }} {{12\sqrt[3]{2}}} + O\left( {\frac{1} {{k^2 }}} \right) $$ To be more clear, you have that $$ \begin{gathered} s_1 \left( k \right) = \frac{1} {{\sqrt[3]{{4k}} + 1}} - \frac{1} {{\sqrt[3]{{4k + 1}} - 1}} \hfill \\ \hfill \\ = \frac{{\sqrt[3]{{4k + 1}} - \sqrt[3]{{4k}} - 2}} {{\left[ {\sqrt[3]{{4k}} + 1} \right]\left[ {\sqrt[3]{{4k + 1}} - 1} \right]}} \hfill \\ \hfill \\ = \frac{{\sqrt[3]{{4k}}\left[ {\sqrt[3]{{1 + \frac{1} {{4k}}}} - 1} \right] - 2}} {{\left[ {\sqrt[3]{{4k}} + 1} \right]\left[ {\sqrt[3]{{4k}}\left( {\sqrt[3]{{1 + \frac{1} {{4k}}}}} \right) - 1} \right]}} \hfill \\ \end{gathered} $$ and then, from here, you can use the Taylor expansion with $$ {\sqrt[3]{{1 + \frac{1} {{4k}}}}} $$

Consider now $$ s_2 \left( k \right) = \frac{1} {{\sqrt[3]{{4k + 2}} - 1}} - \frac{1} {{\sqrt[3]{{4k + 3}} + 1}} $$ and repeat the above procedure. You will get $$ s_2 \left( k \right) = \frac{{\left( {\frac{1} {k}} \right)^{2/3} }} {{\sqrt[3]{2}}} + \frac{{7\left( {\frac{1} {k}} \right)^{4/3} }} {{12\;2^{2/3} }} - \frac{{5\left( {\frac{1} {k}} \right)^{5/3} }} {{12\sqrt[3]{2}}} + O\left( {\frac{1} {{k^2 }}} \right) $$ Therefore $$ s\left( k \right) = s_1 \left( k \right) + s_2 \left( k \right) = \frac{{\left( {\frac{1} {k}} \right)^{4/3} }} {{6 \cdot \;2^{2/3} }} - \frac{{\left( {\frac{1} {k}} \right)^{5/3} }} {{3\sqrt[3]{2}}} + O\left( {\frac{1} {{k^2 }}} \right) $$ so that $$ s\left( k \right) = s_1 \left( k \right) + s_2 \left( k \right) = \frac{A} {{k^{4/3} }} + O\left( {\frac{1} {{k^{5/3} }}} \right) $$ and the considered series is convergent as well as yours.

By the way: the above expansions which contain non integer powers of $\frac{1}{k}$ are called Puiseux expansions.