Series JEE

Algebra Level 5

∑ r = 1 ∞ r 3 + ( r 2 + 1 ) 2 r 6 + r 5 + r 4 + r 3 + r 2 + r \displaystyle \sum_{r=1}^{\infty}\frac{r^{3}+(r^{2}+1)^{2}}{r^{6}+r^{5}+r^{4}+r^{3}+r^{2}+r}

Evaluate the above expression.

This problem is a part of my set Some JEE problems


The answer is 1.50.

View wiki
Tanishq Varshney by Tanishq Varshney , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Tanishq Varshney
Feb 16, 2015

∑ r = 1 ∞ r 3 + 2 r 2 + 1 + r 4 ( r 2 + r ) ( r 4 + r 2 + 1 ) \displaystyle \sum_{r=1}^{\infty}\frac{r^{3}+2r^{2}+1+r^{4}}{(r^{2}+r)(r^{4}+r^{2}+1)}

∑ r = 1 ∞ r 4 + r 2 + 1 + r 2 ( r + 1 ) r ( r + 1 ) ( r 4 + r 2 + 1 ) \displaystyle \sum_{r=1}^{\infty}\frac{r^{4}+r^{2}+1+r^{2}(r+1)}{r(r+1)(r^{4}+r^{2}+1)}

∑ r = 1 ∞ 1 r ( r + 1 ) + ∑ r = 1 ∞ r ( r 4 + r 2 + 1 ) \displaystyle \sum_{r=1}^{\infty}\frac{1}{r(r+1)}+\sum_{r=1}^{\infty}\frac{r}{(r^{4}+r^{2}+1)}

lim ⁡ n → ∞ ( ∑ r = 1 n ( r + 1 ) − r r ( r + 1 ) + 1 2 ∑ r = 1 n 2 r ( r 2 + r + 1 ) ( r 2 − r + 1 ) ) \displaystyle \lim_{n\to \infty}(\sum_{r=1}^{n}\frac{(r+1)-r}{r(r+1)}+\frac{1}{2}\sum_{r=1}^{n}\frac{2r}{(r^{2}+r+1)(r^{2}-r+1)})

lim ⁡ n → ∞ ( ( ∑ r = 1 n 1 r − 1 r + 1 ) + 1 2 ( ∑ r = 1 n 1 r 2 − r + 1 − 1 r 2 + r + 1 ) ) \displaystyle \lim_{n\to \infty}((\sum_{r=1}^{n}\frac{1}{r}-\frac{1}{r+1}) +\frac{1}{2}(\sum_{r=1}^{n}\frac{1}{r^{2}-r+1}-\frac{1}{r^{2}+r+1}))

lim ⁡ n → ∞ ( 1 − 1 n + 1 + 1 2 ( 1 − 1 n 2 + n + 1 ) ) \displaystyle \lim_{n\to \infty}(1-\frac{1}{n+1}+\frac{1}{2}(1-\frac{1}{n^{2}+n+1}))

lim ⁡ n → ∞ ( n n + 1 + 1 2 ( n 2 + n n 2 + n + 1 ) ) \displaystyle \lim_{n\to \infty}(\frac{n}{n+1}+\frac{1}{2}(\frac{n^{2}+n}{n^{2}+n+1}))

[ Apply L hospital rule as it is ∞ ∞ \frac{\infty}{\infty} form ]

1 + 1 2 = 3 2 1+\frac{1}{2}=\boxed{\frac{3}{2}}

Do upvote

21 Helpful 0 Interesting 1 Brilliant 0 Confused

Don't always apply L.H

lim ⁡ n → ∞ ( n n + 1 + 1 2 ( n 2 + n n 2 + n + 1 ) ) \displaystyle \lim_{n\to \infty}(\frac{n}{n+1}+\frac{1}{2}(\frac{n^{2}+n}{n^{2}+n+1}))

lim ⁡ n → ∞ ( 1 1 + 1 n + 1 2 ( 1 + 1 n 1 + 1 n + 1 n 2 ) ) \displaystyle \lim_{n\to \infty}(\dfrac{1}{1+\dfrac{1}{n}}+\frac{1}{2}(\frac{1+\dfrac{1}{n}}{1 +\dfrac{1}{n} +\dfrac{1}{n^2}}))

= 3 2 = \dfrac{3}{2}

U Z - 6 years, 3 months ago

Log in to reply

i would have written the same but i was tired of writing a long solution

Tanishq Varshney - 6 years, 3 months ago

I like your tricks very much!

Nihar Mahajan - 6 years, 2 months ago
Aareyan Manzoor
Feb 18, 2015

first factor = r 4 + r 3 + 2 r 2 + 1 r 6 + r 5 + r 4 + r 3 + r 2 + r = r 5 + r 4 + r 3 + r 2 + r + 1 r 6 + r 5 + r 4 + r 3 + r 2 + r − r 5 − r 2 + r r 6 + r 5 + r 4 + r 3 + r 2 + r = 1 r − r 4 − r + 1 r 5 + r 4 + r 3 + r 2 + r + 1 = 1 r − r 4 − r + 1 ( r + 1 ) ( r 4 + r 2 + 1 ) = 1 r − r 4 + r 2 + 1 − r 2 − r ( r + 1 ) ( r 4 + r 2 + 1 ) = 1 r − 1 r + 1 + r r 4 + r 2 + 1 = 1 r − 1 r + 1 + 1 2 ( ( r 2 + r + 1 ) − ( r 2 − r + 1 ) ( r 2 + r + 1 ) ( r 2 − r + 1 ) ) = 1 r − 1 r + 1 + 1 2 ( 1 r 2 − r + 1 − 1 r 2 + r + 1 ) \begin{array}{c}=&\dfrac{r^4+r^3+2r^2+1}{r^6+r^5+r^4+r^3+r^2+r}\\ =&\dfrac{r^5+r^4+r^3+r^2+r+1}{r^6+r^5+r^4+r^3+r^2+r}-\dfrac{r^5-r^2+r}{r^6+r^5+r^4+r^3+r^2+r}\\ =&\dfrac{1}{r}-\dfrac{r^4-r+1}{r^5+r^4+r^3+r^2+r+1}\\ =&\dfrac{1}{r}-\dfrac{r^4-r+1}{(r+1)(r^4+r^2+1)}\\ =&\dfrac{1}{r}-\dfrac{r^4+r^2+1-r^2-r}{(r+1)(r^4+r^2+1)}\\ =&\dfrac{1}{r}-\dfrac{1}{r+1}+\dfrac{r}{r^4+r^2+1}\\ =&\dfrac{1}{r}-\dfrac{1}{r+1}+\dfrac{1}{2}(\dfrac{(r^2+r+1)-(r^2-r+1)}{(r^2+r+1)(r^2-r+1)})\\ =&\dfrac{1}{r}-\dfrac{1}{r+1}+\dfrac{1}{2}(\dfrac{1}{r^2-r+1}-\dfrac{1}{r^2+r+1}) \end{array} hence, ∑ r = 1 ∞ ( r 4 + r 3 + 2 r 2 + 1 r 6 + r 5 + r 4 + r 3 + r 2 + r ) = \sum_{r=1}^\infty(\dfrac{r^4+r^3+2r^2+1}{r^6+r^5+r^4+r^3+r^2+r})= ∑ r = 1 ∞ ( 1 r − 1 r + 1 ) + 1 2 ∑ r = 1 ∞ ( 1 r 2 − r + 1 − 1 r 2 + r + 1 ) \sum_{r=1}^\infty(\dfrac{1}{r}-\dfrac{1}{r+1})+\dfrac{1}{2}\sum_{r=1}^\infty(\dfrac{1}{r^2-r+1}-\dfrac{1}{r^2+r+1}) both are telescoping series which result in 1 1 + 1 2 ∗ 1 = 3 2 ≈ 1.5 1+\dfrac{1}{2}*1=\dfrac{3}{2}\approx 1.5

7 Helpful 0 Interesting 1 Brilliant 0 Confused

same thing that I did because I didn't know what was L'hopital rule.

Nihar Mahajan - 6 years, 2 months ago

same thing that I did. Doesn't require L'hopital rule

Anirban Ghosh - 6 years, 3 months ago
Khurram Sami
Feb 22, 2015

Yep. Same thing . Doesn't require L hopital rule

0 Helpful 0 Interesting 0 Brilliant 0 Confused

1 pending report

Vote up reports you agree with

×

Problem Loading...

Note Loading...

Set Loading...