13th Harvard MIT Calculus problem 8


Let f(n)=\sum\limits_{k=2}^{\infty }{\frac{1}{{{k}^{n}}k!}}. Calculate \sum\limits_{k=2}^{\infty }{f(n)} 😀

Solution

—> \sum\limits_{k=2}^{\infty }{f(n)}=\sum\limits_{k=2}^{\infty }{\sum\limits_{n=2}^{\infty }{\frac{1}{{{k}^{n}}k!}}}

=\sum\limits_{k=2}^{\infty }{\frac{1}{k!}\sum\limits_{n=2}^{\infty }{\frac{1}{{{k}^{n}}}}}

we know that \sum\limits_{n=0}^{\infty }{\frac{1}{{{k}^{n}}}}=\frac{1}{{{k}^{0}}}+\frac{1}{{{k}^{1}}}+...........=\frac{1}{1-\frac{1}{k}}=\frac{k}{k-1}

So \sum\limits_{n=2}^{\infty }{\frac{1}{{{k}^{n}}}}=\frac{k}{k-1}-1-\frac{1}{k}=\frac{1}{k(k-1)}

—> \sum\limits_{n=2}^{\infty }{f(n)}=\sum\limits_{k=2}^{\infty }{\frac{1}{k!}.\frac{1}{k(k-1)}}

=\sum\limits_{k=2}^{\infty }{\frac{1}{(k-1)!}}.\frac{1}{{{k}^{2}}(k-1)}

–>  =\sum\limits_{k=2}^{\infty }{\frac{1}{(k-1)!}\left( \frac{1}{k-1}-\frac{1}{{{k}^{2}}}-\frac{1}{k} \right)}

=\sum\limits_{k=2}^{\infty }{\left( \frac{1}{(k-1)!(k-1)}-\frac{1}{k.k!}-\frac{1}{k!} \right)}

=\sum\limits_{k=2}^{\infty }{\left( \frac{1}{(k-1)!(k-1)}-\frac{1}{k.k!} \right)}-\sum\limits_{k=2}^{\infty }{\frac{1}{k!}}

*** By Taylor series \sum\limits_{k=0}^{\infty }{\frac{1}{k!}}=e

Therefore \sum\limits_{n=2}^{\infty }{f(n)=\frac{1}{1\cdot 1!}-\left( e-\frac{1}{0!}-\frac{1}{1!} \right)=3-e}

 

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About Bunchhieng

I like math, programming and web design!!
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