Question

Compute the sum and the limit of the sum as n approaches infinity.

Answer 1

@satellite73 can u help?

Answer 2

yeah we can do this

Answer 3

\[\sum_{i=1}^{\infty}\frac{1}{n}[\frac{i^2}{n^2}+\frac{i}{n}]\] first of don't get confused by the \(n\) in the denominator, the index is \(i\) so we can bring the \(n\) right out front of the summation

Answer 4

okay

Answer 5

first rewrite as
\[\frac{1}{n^3}\sum i^2+\frac{i}{n^2}\sum i\]

Answer 6

ok

Answer 7

with this so far?

Answer 8

yes

Answer 9

now we use the summation formulas for
\[\sum i^2\] and \[\sum i\]

Answer 10

okay

Answer 11

\[\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6}\] and
\[\sum_{i=1}^ni=\frac{n(n+1)}{2}\] two more or less well known formulas

Answer 12

okay i see

Answer 13

so the first term is
\[\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6}\] and the second term is
\[\frac{1}{n^2}\frac{n(n+1)}{2}\]

Answer 14

ok

Answer 15

so far so good?

Answer 16

yea

Answer 17

your last job is to take the limit as \(n\to \infty\) of each term, which you can do more or less by your eyeballs

Answer 18

okau thx

Answer 19

i hope the last step is clear, and you don't do much work at all

Answer 20

for example
\[\frac{1}{n^2}\frac{n(n+1)}{2}\] is a rational function with numerator and denominator degree 2, to the limit is the ratio of the leading coefficients, namely \(\frac{1}{2}\)

Answer 21

okay i get it

Answer 22

good! not too bad, was it ?

Answer 23

nope thanks!