infinite series - 9

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \sum_{k=1}^{\infty}(-1)^k\frac{2}{k^2}\)

 

[khansaheb]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \sum_{k=1}^{\infty}(-1)^k\frac{2}{k^2}\)

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[logistic_guy]

\(\displaystyle \sum_{k=1}^{\infty}(-1)^k\frac{2}{k^2}\)

Apply the absolute value.

\(\displaystyle \sum_{k=1}^{\infty}\left|(-1)^k\frac{2}{k^2}\right| = \sum_{k=1}^{\infty}\frac{2}{k^2}\)

We have a beautiful theorem which says:

If \(\displaystyle \sum_{k=1}^{\infty}\frac{2}{k^2}\) converges, then

\(\displaystyle \sum_{k=1}^{\infty}(-1)^k\frac{2}{k^2}\) converges as well.

From the previous post, we know that the series \(\displaystyle \sum_{k=1}^{\infty}\frac{2}{k^2}\) converges by the p-series test since \(\displaystyle 2 > 1\).

Then,

\(\displaystyle \sum_{k=1}^{\infty}(-1)^k\frac{2}{k^2}\) converges by the absolute convergence test.

Or

\(\displaystyle \sum_{k=1}^{\infty}(-1)^k\frac{2}{k^2} \longrightarrow \textcolor{green}{\bold{converges \ absolutely}}\).