How to rewrite a Sum?

Question

How to find the following sum? $$ \frac{2}{n} \sum_{i=1}^{n} \left(\frac{4i}{n} +1\right) $$ For example $$ \sum_{i = 1}^n i^2 = \frac{n(n+1)(2n+1)}{6}. $$ I need to find the value of which this would converge to and I know that it must be six because the integral from $1$ to $3$ of the function $(2x-1)$ is $6$.

Any help is appreciated!!!

A couple of hints: A constant can always be moved in front of the summation, like this: $$ \sum_n {a f(n)} = a \sum_n f(n) $$ And the sum is also distributive: $$ \sum_{n} (a_n + b_n) = \sum_n a_n + \sum_n b_n $$ — Matti P., May 15 '18 at 05:38

Mundron Schmidt · Answer 1 · 2018-05-15T05:45:18.177

9

Hint: You have to use $$\sum_{i=1}^n (a_i+b_i) = \left(\sum_{i=1}^na_i\right)+\left(\sum_{i=1}^n b_i\right)$$ and $$ \sum_{i=1}^n (\lambda a_i)=\lambda\left(\sum_{i=1}^n a_i\right) $$ For the second equation, you have to check that $\lambda$ can not depend on $i$ but it can depend on other variables like $n$.

edited May 15 '18 at 05:45

answered May 15 '18 at 05:39

Mundron Schmidt

7,993

Mundron - Why can we directly split this sum, for infinite case its not allowed right – King Tut May 15 '18 at 05:47
It comes from the associative and distributive law. But in general, you can't use it for infinite sums. – Mundron Schmidt May 15 '18 at 06:14

trancelocation · Answer 2 · 2018-05-15T09:25:51.063

2

$$\frac{2}{n} \sum_{i=1}^{n} \left(\frac{4i}{n} +1\right) = \frac{2}{n}\left( \sum_{i=1}^{n} \frac{4i}{n} + \sum_{i=1}^{n} 1\right) = \frac{8}{n^2}\sum_{i=1}^{n} i + \frac{2}{n}\cdot n = \frac{8}{n^2}\cdot \frac{n(n+1)}{2} + 2 = 4 \left(1+\frac{1}{n} \right) + 2$$

edited May 15 '18 at 09:25

answered May 15 '18 at 06:08

trancelocation

32,243

The final result could have been simplified. – Stephan May 15 '18 at 08:58
4

I wanted to put at least one little challenge to the OP. :-) – trancelocation May 15 '18 at 09:06

How to rewrite a Sum?

2 Answers2