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geekotakuglh

Answered question

2022-02-24

Evaluate the series:

\sum_{k = 1}^{\infty} \frac{1}{k {(k + 1)}^{2} k!}

Abbey Hope

Beginner2022-02-25Added 7 answers

Try it with factor

x^{k}

f (x) = \sum_{k = 1}^{\infty} \frac{x^{k}}{k {(k + 1)}^{2} k!}

f^{'} (x) = \sum_{k = 1}^{\infty} \frac{x^{k - 1}}{{(k + 1)}^{2} k!}

x^{2} f^{'} (x) = \sum_{k = 1}^{\infty} \frac{x^{k + 1}}{{(k + 1)}^{2} k!}

{(x^{2} f^{'} (x))}^{'} = \sum_{k = 1}^{\infty} \frac{x^{k}}{(k + 1) k!}

x {(x^{2} f^{'} (x))}^{'} = \sum_{k = 1}^{\infty} \frac{x^{k + 1}}{(k + 1) k!}

{(x {(x^{2} f^{'} (x))}^{'})}^{'} = \sum_{k = 1}^{\infty} \frac{x^{k}}{k!} = e^{x} - 1

Now solve a differential equation. Then plug in

x = 1

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