Question

limit n->infinity
(1^k +2^k +3^k ....... n^k)/k(n^(k+1))

Answer 1

could you clarify denominator??

Answer 2

\[\lim_{n \rightarrow infinity} (1^k +2^k ...... n^k)/(kn^{k+1})\]

Answer 3

k times n^(k+1)??

Answer 4

yes..

Answer 5

\[ \lim_{n \rightarrow \infty } {1 \over kn^{k + 1}}{\sum_{n=0}^\infty n^k}\]

Answer 6

i guess we have to get the general formula for n^k

Answer 7

is there any??
i also thought of fancy things ,wolframed it,,but it said something about reiman zeta function..
but this is a simple 12th grade problem..

Answer 8

i somehow arrived at the ans by hit and trial by putting k=1,then 2 and 3 ..but thats cheating :|

Answer 9

\[ \sum n^k = \sum (n+1)^{k}\]
i'm not sure if we could use this technique .. looks lie long way to generalize it.
can you post the wolf link?

Answer 10

http://www.wolframalpha.com/input/?i=+%28summation%28n%5Ek%29++from+n%3D1+to+infinity%29%2F%28k%28%28limit+n+tends+to+infinity+n%5E%28k%2B1%29%29%29%29

Answer 11

maybe we can make that substitution, but any progress?

Answer 12

with that i mean **

Answer 13

i guess that incorrect expression ... should be something like this
http://www.wolframalpha.com/input/?i=+lim+n-%3Einf+sum%5Bi%5Ek%2C0%2Cn+%5D%2F%28kn%5E%28k%2B1%29%29
though wolf doesn't take it ... let me check it in mathematica first.

Answer 14

Limit[Sum[i^k, {i, 0, n}]/(k n^(k + 1)), n -> Infinity]
\[ = \text{Limit}\left[\frac{n^{-1-k} \left(0^k+\text{HarmonicNumber}[n,-k]\right)}{k},n\to \infty \right] \]

Answer 15

\( \huge \lim_{n \rightarrow \infty} \frac{1^k+2^k+...+n^k}{k n^{k+1}}\\ \huge=\lim_{n \rightarrow \infty}\frac{1}{k} [(\frac{1}{n})^k+(\frac{2}{n})^k+...+(\frac{n}{n})^k] \frac{1}{n}\\ \huge \frac{1}{k} \int_{0}^{1} x^k dx=\frac{1}{k(k+1)} \)

Answer 16

seems like i've been misinterpreting it whole time.

Answer 17

i was wondering what u did that ;)

Answer 18

reading 1^k + 2^k + 3^k ... as (1 + 2 + 3 + .. )^k lol

Answer 19

lol :)

Answer 20

anyway very nice and beautiful technique ... i haven't seen that before.

Answer 21

i would never have never seen Riemann sums in that series.

Answer 22

excellent @mukushla ..thank you very much :)