0

Let $$\sum_{k=0}^{n-1} \frac{nk^2-2k^3-7kn^2+10n^3}{n^4}=a_n$$ and $$\sum_{k=1}^n \frac{nk^2-2k^3-7kn^2+10n^3}{n^4}=b_n$$ then for $n=1,2,3,4,...$ then question is to find which among following is correct

A)$a_n>19/3$

B)$b_n<19/3$

C)$b_n>19/3$

D)$a_n<19/3$

I tried to bring the summation in some form of telescoping series so that I can get the options checked by putting $n=\infty$ but the fourth power in denominator is causing trouble.Any ideas?Thanks.

Navin
  • 2,605
  • Might be helpful to note that $b-a = f(n) - f(0)$ where $f$ is your summand. – Zain Patel May 08 '17 at 19:16
  • for n=1,2,3,4,... It can't hold for all $n$ since, for example, $b-a=2/n-10$ depends on $n$. – dxiv May 08 '17 at 19:16
  • @dxiv I don't think that since $b-a$ depends on $n$ the summation is not defined for all $n$.The question is to find out extrema of summation. – Navin May 08 '17 at 19:26
  • The summation is defined for all $n$ but the partial sums are not constants. The question would be more clear if you used $a_n,b_n$ instead of $a,b$. – dxiv May 08 '17 at 19:29
  • @dxiv thanks for pointing out. – Navin May 08 '17 at 19:32

1 Answers1

3

By brute force: $$ \begin{align} a_n &= \sum_{k=0}^{n-1} \frac{nk^2-2k^3-7kn^2+10n^3}{n^4} \\ &= \frac{1}{n^3} \sum_{k=0}^{n-1} k^2 - \frac{2}{n^4} \sum_{k=0}^{n-1} k^3 - \frac{7}{n^2} \sum_{k=0}^{n-1} k + \frac{10}{n} \sum_{k=0}^{n-1} 1 \\ &= \frac{1}{n^3} \cdot \frac{n(n-1)(2n-1)}{6}- \frac{2}{n^4} \cdot \frac{n^2(n-1)^2}{4} - \frac{7}{n^2} \cdot \frac{n(n-1)}{2} + \frac{10}{n}\cdot n \\ &= \frac{2(n-1)(2n-1) - 6(n-1)^2-42n(n-1)+120n^2}{12 n^2} \\ &= \frac{4 (19 n^2 + 12 n - 1)}{12n^2} \\[3px] &= \frac{19}{3}+\frac{12n-1}{3n^2} \;\;\gt\;\; \frac{19}{3} \end{align} $$

dxiv
  • 76,497
  • Thanks for your answer.this basically tells us that $a_n \geq 2$ and so is $b_n$.But it does not tell whether $a_n$ or $b_n$ is greater than $19/3$ or not. – Navin May 08 '17 at 19:45
  • @Pink Right, previous hint retracted. The reworked answer may not be pretty but should do the job. – dxiv May 08 '17 at 20:22
  • Thanks again.using your expression I noticed that the value of $a_n$ equal 19/3 in limit $n \to \infty$ which indicates there might be something more simple way to do it. – Navin May 08 '17 at 20:34
  • @Pink That would be Riemann sums, if you know integrals:

    $$ b_n ,\lt, \int_0^1 (x^2 - 2 x^3 - 7 x + 10) ,dx = \frac{19}{3} ,\lt, a_n $$

     – dxiv May 08 '17 at 20:40
  • Yes I get it. by pulling out $n$ from denominator $a_n$ can be converted into reimann sum by taking limit $n \to \infty$that's the simplest to evaluate...But how did you get that $b_n<a_n$ using reimann sum.I thought that both are equal to each other as $n \to \infty$ – Navin May 08 '17 at 20:48
  • 1
    @Pink The integrand is decreasing on $[0,1]$ and $b_n$ is the approximation from below, $a_n$ from above. In the limit $n \to \infty$ yes, both $a_n$ and $b_n$ tend to the same value, which is the value of the integral. – dxiv May 08 '17 at 20:57