infinite series - 10

[logistic_guy]

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

\(\displaystyle \sum_{k=0}^{\infty}(-1)^k\frac{6}{k!}\)

 

[khansaheb]

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

\(\displaystyle \sum_{k=0}^{\infty}(-1)^k\frac{6}{k!}\)

Please show us what you have tried and exactly where you are stuck.

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[logistic_guy]

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

\(\displaystyle \sum_{k=0}^{\infty}(-1)^k\frac{6}{k!}\)

What about using the ratio test?

\(\displaystyle \lim_{k\rightarrow \infty}\frac{6}{(k+1)!}\frac{k!}{6} = \lim_{k\rightarrow \infty}\frac{6}{(k+1)k!}\frac{k!}{6} = \lim_{k\rightarrow \infty}\frac{1}{k+1} = 0 < 1\)

Then,

the infinite series \(\displaystyle \sum_{k=0}^{\infty}(-1)^k\frac{6}{k!}\) converges by the ratio test.

Or

\(\displaystyle \sum_{k=0}^{\infty}(-1)^k\frac{6}{k!} \ \textcolor{indigo}{\text{converges absolutely.}}\)