How to evaluate this limit when n tends to +infinity
$(1^4+2^4+...n^4)/n^5 - (1^3+2^3+...+n^3)/n^5$
I tried using L hopital but got answer zero which is incorrect, then I tried to somehow use some algebraic identity but couldn't figure out one.
Asked
Active
Viewed 81 times
-1
-
Used the formalue for sum of cubes and fourth powers? – Juanito May 14 '17 at 05:21
-
shouldn't you get 1/5 by L'hopital????? – Saketh Malyala May 14 '17 at 05:21
-
Is there some other method because The sum of fourth powers is little complex and will make it lengthy ? – Matt May 14 '17 at 05:22
-
@SakethMalyala how when we will take derivative in both cases above will be 4n^3 and 3n^2 smaller than below power hence zero. – Matt May 14 '17 at 05:23
-
Use https://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2/469886#469886 – lab bhattacharjee May 14 '17 at 05:23
-
Okay. Thanks can we do this purely by limits not by integration. – Matt May 14 '17 at 05:24
-
Last $n^5$ should read $n^4$? – Did May 14 '17 at 07:03
1 Answers
1
METHOD 1. (Long. No calculus): For integer $j$ let $S(n,j)=(1^j+...+n^j)$ and let $T(n,j)=S(n,j)/n^5.$ You are asking for $\lim_{n\to \infty}T(n,4)-T(n,3).$
(1)... If $j\leq 3$ then $\lim_{n\to \infty}T(n,j)=0.$ Proof: $$0<T(n,j)=(1^j+...+n^j)/n^5\leq (n^j+...+n^j)/n^5=(n\cdot n^j)/n^5\leq 1/n. $$
(2)... We have $(x+1)^5-x^5=5x^4+10x^3+10x^2+5x+1.$ Summing this from $x=1$ to $x=n $ we have $$ (i) \quad \sum_{x=1}^n(x+1)^5-x^5=5S(n,4)+10S(n,3)+10S(n,2)+5S(n,1)+S(n,0).$$ The LHS of $(i)$ is a telescoping sum, and is equal to $(n+1)^5-1.$ For example, when $n=3$ we have $(4^5-3^5)+(3^5-2^5)+(2^5-1^5)=4^5-1.$ So, dividing both sides of $(i)$ by $n^5$ we have $$ (ii) \quad \left(1+\frac {1}{n}\right)^5-n^{-5}=5T(n,4)+10T(n,3)+10T(n,2)+5T(n,1)+T(n,0).$$ The LHS of $(ii)$ tends to $1$ as $n\to \infty.$ And we have seen that $T(n,j)\to 0$ as $n\to \infty$ when $j\in \{0,1,2,3\}$. Therefore $$1=\lim_{n\to \infty}5T(n,4).$$ $$ \text {Therefore }\quad \lim_{n\to \infty}T(n,4)-T(n,3)=\lim_{n\to \infty}T(n,4)=1/5.$$
METHOD 2: Let $j\geq 0.$ The function $f(t)=t^j$ is increasing for $t>0.$ So for $x\geq 1$ we have $$\int_{x-1}^x t^jdt\leq x^j\leq \int_x^{x+1}t^jdt.$$ Summing this inequality from $x=1$ to $x=n$ gives $$ n^{j+1}/(j+1)= \int_0^nt^jdt\leq \;(1^j+...+n^j)\;\leq \int_1^{n+1}t^jdt=((n+1)^{j+1}-1)/(j+1).$$ And the rest is easy.
DanielWainfleet
- 57,985