# What is sum_(k=0)^(4n+2)1/(k+1)+1/(k+3) ?

##### 1 Answer
Jun 20, 2018

$S = 2 {H}_{4 n + 3} + \frac{1}{4 n + 4} + \frac{1}{4 n + 5} - \frac{3}{2}$

#### Explanation:

We seek a way to evaluate the sum

$S = {\sum}_{k = 0}^{m} \left(\frac{1}{k + 1} + \frac{1}{k + 3}\right)$

Note:

• The substitution color(red)(m=4n+2, is made for simplicity

Definition:

The $n$-th harmonic number ${H}_{n}$ is a number on the form

color(blue)(H_n=sum_(k=1)^(n)1/k

Our goal will be to express the original sum in terms of a harmonic number.

It may seems strange to express the original sum, by another sum. But, you may think of the harmonic numbers, similar to the factorial, in the way evaluate them.

Express the sum in terms of a harmonic number:

Using some basic summation identities

$S = {\sum}_{k = 0}^{m} \frac{1}{k + 1} + {\sum}_{k = 0}^{m} \frac{1}{k + 3}$

$\textcolor{w h i t e}{S} = {\sum}_{k = 1}^{m + 1} \frac{1}{k} + {\sum}_{k = 3}^{m + 3} \frac{1}{k}$

$\textcolor{w h i t e}{S} = {\sum}_{k = 1}^{m + 1} \frac{1}{k} + {\sum}_{k = 1}^{m + 3} \frac{1}{k} - \frac{3}{2}$

$\textcolor{w h i t e}{S} = 2 {\sum}_{k = 1}^{m + 1} \frac{1}{k} + \frac{1}{m + 2} + \frac{1}{m + 3} - \frac{3}{2}$

Or in terms of a harmonic number

$S = 2 {H}_{m + 1} + \frac{1}{m + 2} + \frac{1}{m + 3} - \frac{3}{2}$

For your problem you may substitute back color(red)(m=4n+2

$S = 2 {H}_{4 n + 3} + \frac{1}{4 n + 4} + \frac{1}{4 n + 5} - \frac{3}{2}$

Bonus info

A fairly reasonable approximation of a harmonic number is

${H}_{n} \approx \ln \left(n\right) - \gamma$

The drawing at the upper right hand corner illustrates this quite well