Sums 1

Calculus Level 3

Find the value of n = 1 6 n n 4 . \displaystyle \sum_{n=1}^\infty\frac {6n}{n^4} .

Give your answer to 1 decimal place.


The answer is 7.2.

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2 solutions

Akshay Yadav
Dec 8, 2015

First of all simplification,

S = n = 1 6 n n 4 S=\displaystyle\sum_{n=1}^{\infty} \frac{6n}{n^{4}}

S = 6 n = 1 1 n 3 S=6\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{3}}

We know,

n = 1 1 n 3 = 1.21 \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{3}}=1.21

Hence,

S = 6 × 1.21 S=6\times1.21

S = 7.26 S=7.26

How you knew that n = 1 1 n 3 = 1.21 \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{3}}=1.21 ?

Akshat Sharda - 5 years, 6 months ago

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We use here the riemman zeta function.

Imad Fatimy - 5 years, 6 months ago
Pulkit Gupta
Dec 7, 2015

The general term is 6 / n 3 n^3 .

Σ \Sigma 1 / n 3 n^3 is ζ \zeta (3) = 1.20 ( approx). The answer is obtained by multiplying this value obtained by 6 which is clear by looking at the general term.

The main point in here is to use the Riemman zeta function. :) Cheers!

Imad Fatimy - 5 years, 6 months ago

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