If $\lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{(n+1)^{a-1}.((na+1)+(na+2)+...+(na+n))}=\frac{1}{60}$, Find the value of a

If $$\lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{(n+1)^{a-1}\cdot((na+1)+(na+2)+...+(na+n))}=\frac{1}{60}$$ Find the value of a.

Attempt: I solved it using two methods each giving me different answers.

$$=\lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{(n+1)^{a-1}\cdot(n^2a+\frac{n(n+1)}{2})}$$ $$=\lim_{n\to\infty}\frac{1^a+2^a+\dots+n^a}{n^{2}(n+1)^{a-1}\cdot(a+\frac{(1+1/n)}{2})}$$$$=2\lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{n^{a+1}(1+1/n)^{a-1}(2a+1+1/n)}$$$$=2\lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{n^{a+1}(2a+1)}=0$$ since degree of denominator is greater than that of numerator. So you wont get 1/60. But if I use method of integration, the given expression can be written as $$\displaystyle\lim_{n\to\infty}\frac{n^a\sum_{r=1}^n(\frac{r}{n})^a}{(n+1)^{a-1}\cdot n\cdot\sum_{r=1}^n(a+\frac{r}{n})}$$$$\displaystyle=\dfrac{\int_0^1x^a\,dx}{\int_0^1(a+x)\,dx}$$$$=\frac{2}{(2a+1)(a+1)}$$ This gives a= 7 or -17/2 which is the right answer. What is wrong with my first method?

Solution 1:

\begin{align} \frac{1}{60}&=\lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{(n+1)^{a-1}.(n^2a+\frac{n(n+1)}{2})}\\ &=\lim_{n\to\infty}\frac{n^{a+1}\frac 1n\sum_{j=1}^n\Big(\frac jn\Big)^a}{(n+1)^{a-1}.(n^2a+\frac{n(n+1)}{2})}\\ &=\lim_{n\to\infty}\frac{n^{a+1}}{(n+1)^{a-1}.(n^2a+\frac{n(n+1)}{2})}\times \lim_{n\to\infty}\frac 1n\sum_{j=1}^n\Big(\frac jn\Big)^a\\ &=\frac{1}{a+\frac12}\int_0^1x^adx\\ &=\frac{1}{a+\frac12}\frac{1}{a+1} \end{align} where the integral requires $a>-1$. Therefore $a=7$ is the only acceptable solution.

Your first method is flawed. The limit is not zero.

Solution 2:

Not sure how to carry method (1) out to find the value of $a,$ but since your question is

What is wrong with my first method?

I will say that step line 4 of that solution is identical to line 2, and line 5 should read (carrying on from line 3 and skipping 4):

$$2\lim_{n\to\infty}\frac{1^a+2^a+\cdots+n^a}{2an^{a+1}(1+1/n)^{a-1}+n^{a+1}(1+1/n)^a}.$$

Solution 3:

In the first attempt, you have taken the highest power of numerator as $a$ but in the summation, the highest power will become $a+1$, and hence the powers for numerator and denominator become the same.