Ferocious Series?

Algebra Level 2

$\frac {\color{#D61F06}{3}} {\color{#69047E}{4}} + \frac {\color{#D61F06}{5}} {\color{#69047E}{36}} + \frac { \color{#D61F06}{7} } {\color{#69047E}{144}} + \frac {\color{#D61F06}{9}}{\color{#69047E}{400}} + \frac {\color{#D61F06}{11}}{\color{#69047E}{900}} + \ldots = \ ?$

Want more of my sets Click here

The answer is 1.000.

5 solutions

Purushottam Abhisheikh
Mar 23, 2015

$S_n = \dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+\dfrac{9}{400}+\dfrac{11}{900}+\dots$

$\implies S_n = \dfrac{3}{2^2}+\dfrac{5}{6^2}+\dfrac{7}{12^2}+\dots$

$\implies S_n = 1 - \dfrac{1}{2^2} +\dfrac{1}{2^2} -\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+\dots$

$\implies S_n = 1$

For those who are wondering what he did in the transition from 2nd step to 3rd step, here is the explanation:

$S_n=\sum_{k=1}^\infty \frac{2k+1}{k^2(k+1)^2}\\ \implies S_n=\sum_{k=1}^\infty \frac{(k+1)^2-k^2}{k^2(k+1)^2}=\sum_{k=1}^\infty \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right)$

As this is a telescoping sum, all the subsequent terms except $\dfrac{1}{1^2}=1$ get cancelled out. Hence, the required value is $\boxed{1}$

P.s - Since the identification of the denominator of the general term is not that obvious, a demonstration can be provided to find the general term that involves recurrence relations. Key point to note is that the denominator terms are perfect squares with the sequence of denominators $\displaystyle\left\{(a_i)^2\right\}_{i=1}^{i=k}$ satisfying the following recurrence:

$a_k-a_{k-1}=2k~,k\geq 2~,~a_1=2$

Solving this gives us the closed form $a_k=k(k+1)~,k\geq 1$ . The general term of the sum is,

$t_k=\frac{2k+1}{(a_k)^2}~,k\geq 1$

Using the solution of the recurrence we just solved, we obtain the general term of the sum and proceed as usual.

Prasun Biswas - 6 years, 2 months ago

Sujoy Roy
Mar 23, 2015

$r-$ th term of the given equation is $t_{r} = \frac{2r+1}{[r(r+1)]^2} = \frac{(r+1)^2-r^2}{[r(r+1)]^2} =\frac{1}{r^2}-\frac{1}{(r+1)^2}$ .

Now $S_{n}=\displaystyle\sum_{r=1}^{n}t_{r}$

$=(1-\frac{1}{2^2})+(\frac{1}{2^2}-\frac{1}{3^2})+\ldots+(\frac{1}{n^2}-\frac{1}{(n+1)^2})=1-\frac{1}{(n+1)^2}$ .

Required sum = $S_{\infty}=\displaystyle\lim_{n\rightarrow \infty} [1-\frac{1}{(n+1)^2}]=\boxed{1}$ .

Otto Bretscher
Mar 24, 2015

Let $S_n=\sum_{k=1}^{n}\frac{2k+1}{k^2(k+1)^2}$ . We will prove by induction that $S_n=1-\frac{1}{(n+1)^2}$ . Indeed, $S_1=\frac{3}{4}$ and $S_n=S_{n-1}+\frac{2n+1}{n^2(n+1)^2}$ $=1-\frac{1}{n^2}+\frac{2n+1}{n^2(n+1)^2}=1-\frac{1}{(n+1)^2}$ . Now $\sum_{k=1}^{\infty}\frac{2k+1}{k^2(k+1)^2}=\lim_{n\to\infty}S_n=1$ .

Gamal Sultan
Mar 24, 2015

3/4 = 1 - 1/4

5/36 = 1/4 - 1/9

7/144 = 1/9 - 1/16

9/400 = 1/16 - 1/25

11/900 = 1/25 - 1/36

............................................ ............................................ etc

Adding

The sum = 1 - (1/4 - 1/4) - (1/9 - 1/9) - (1/16 - 1/16) - (1/25 - 1/25) - ... ..........

= 1 - 0 - 0 - 0 -..................... = 1

Worranat Pakornrat
Mar 24, 2015

Write a solution. The nth term of the series can be expressed as (2n+1)/[(n^2)(n+1)^2].

So (2n+1)/[(n^2)(n+1)^2] = n/[(n^2)(n+1)^2] + (n+1)/[(n^2)(n+1)^2]

  = 1/(n(n+1)^2) + 1/[(n^2)(n+1)]

  =  (n+1-n)/(n(n+1)^2)  + (n+1-n)/[(n^2)(n+1)]

  = (n+1)/(n(n+1)^2) - n/(n(n+1)^2) + (n+1)/[(n^2)(n+1)] - n/[(n^2)(n+1)]

  = 1/(n(n+1)) - 1/((n+1)^2) + 1/(n^2) - 1/(n(n+1)) 

  =  1/(n^2)- 1/((n+1)^2)

Sum of 1/(n^2) = S1 = 1/1 + 1/(2^2) + 1/(3^2) + 1/(4^2)+...

Sum of 1/((n+1)^2) = S2 = 1/(2^2) + 1/(3^2) + 1/(4^2)+...

Thus, S1-S2 = 1. The sum of the series is, therefore, equal to 1.

Ferocious Series?

The answer is 1.000.

5 solutions

0 pending reports