Inspired by MIT problem (I)

Calculus Level 3

f ( r ) = ∑ n = 2 2018 1 n r = 1 2 r + 1 3 r + ⋯ + 1 201 8 r f(r)=\sum_{n=2}^{ 2018} \frac{1}{n^r}= \frac{1}{2^r}+\frac{1}{3^r}+\dots+\frac{1}{2018^r}

For f ( r ) f(r) as defined above, find ∑ k = 2 ∞ f ( k ) \displaystyle \sum_{k=2}^{\infty}f(k) .

2018 2019 \frac{2018}{2019} 2015 2016 \frac{2015}{2016} 2016 2017 \frac{2016}{2017} 2019 2020 \frac{2019}{2020} 2017 2018 \frac{2017}{2018}

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

S = ∑ k = 2 ∞ ∑ n = 2 2018 1 n k = ∑ n = 2 2018 ∑ k = 2 ∞ 1 n k = ∑ n = 2 2018 1 n 2 ( 1 1 − 1 n ) = ∑ n = 2 2018 1 n ( n − 1 ) = ∑ n = 2 2018 ( 1 n − 1 − 1 n ) = 1 1 − 1 2018 = 2017 2018 \begin{aligned} S & = \sum_{k=2}^\infty \sum_{n=2}^{2018} \frac 1{n^k} \\ & = \sum_{n=2}^{2018} \sum_{k=2}^\infty \frac 1{n^k} \\ & = \sum_{n=2}^{2018} \frac 1{n^2} \left(\frac 1{1-\frac 1n}\right) \\ & = \sum_{n=2}^{2018} \frac 1{n(n-1)} \\ & = \sum_{n=2}^{2018} \left(\frac 1{n-1} - \frac 1n\right) \\ & = \frac 11 - \frac 1{2018} \\ & = \boxed{\dfrac {2017}{2018}} \end{aligned}

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Thank you Sir.

Hana Wehbi - 2 years, 8 months ago

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