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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?

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  • $\begingroup$ I am not certain of this, but it might have to do with the statement that the denominator has higher degree than the numerator. If the numerator were a sum of $n$'s to powers, it would be the case, but the numerator is not such a sum. If you take a small value of $a$ and check this yourself, it should be clear that the numerator has a degree greater than or equal to the denominator. $\endgroup$
    – Terra Hyde
    Jul 22, 2015 at 17:54
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    $\begingroup$ In the first method, the "degree of numerator" is not $a$ since the number of terms in the numerator increases with $n$. If such a degree had to be defined at all cost, it would rather be $a+1$, at least when $a>-1$. Motivation when $a>0$: there are $n$ terms at most $n^a$ hence the numerator is at most $n^{a+1}$, and there are $n/2$ terms at least $(n/2)^a$ hence the numerator is at least $n^{a+1}/2^a$. $\endgroup$
    – Did
    Jul 22, 2015 at 17:55

5 Answers 5

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\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.

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    $\begingroup$ ,i do not understand why does integral require $a>-1$.Rest of the question is clear to me. $\endgroup$ Feb 2, 2016 at 14:55
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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):

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

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  • $\begingroup$ I am sorry. I did some copy pasting which went wrong. $\endgroup$
    – Aditya Dev
    Jul 22, 2015 at 18:24
  • $\begingroup$ @AdityaDev, does this help you find your answer? $\endgroup$ Jul 22, 2015 at 18:26
  • $\begingroup$ In line 3, all 1/n's will become zero since n tends to infinity. Also, in your expression, if you take away 1/n, you will get the same expression as my line 5. $\endgroup$
    – Aditya Dev
    Jul 22, 2015 at 18:31
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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.

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This gives a= 7 or -17/2 which is the right answer?

For $a = -\frac{17}{2}$ the integral in the numerator does not converge so the answer is $a = 7$.

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  • $\begingroup$ What do you mean by converge? $\endgroup$
    – Aditya Dev
    Jul 22, 2015 at 18:33
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    $\begingroup$ @AdityaDev The integrant of $\int_0^1 \frac{1}{ x^ { \frac{17}{2} } }$ has a sigularity at $x=0$, and as a Riemann integral You can't assign a value to it. Or in other words the Surface under the integrant is infinite. $\endgroup$
    – john
    Jul 22, 2015 at 18:38
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Your method is correct only. (from the last line of your method): If you expand the term n^(a+1) as n.n^a and write the general term of the sum as Σ (r/n^a).1/n and then integrate it, you will get the same answer.

And I don't understand what you mean by "since degree of denominator is greater than that of the numerator. So you wont get 1/60"

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