Telescopic Series

Calculus Level 4

∑ n = 1 ∞ 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) = a b \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (n+1)(n+2)(n+3)(n+4) } } =\frac { a }{ b } a a and b b are co-prime integers. Find a + b a+b .


The answer is 73.

Bhargav Upadhyay by Bhargav Upadhyay , 29, India

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2 solutions

Discussions for this problem are now closed

Kuang-Lin Pan
Feb 28, 2015

Inspired by the way of canceling terms: ∑ n = 1 ∞ 1 ( n + 1 ) ( n + 2 ) = ∑ n = 1 ∞ ( 1 n + 1 − 1 n + 2 ) = ( 1 2 − 2 3 ) + ( 2 3 − 3 4 ) + ( 3 4 − 4 5 ) + … = 1 2 , \sum_{n=1}^\infty \frac{1}{(n+1)(n+2)}\\ =\sum_{n=1}^\infty \left(\frac{1}{n+1}-\frac{1}{n+2}\right)\\ =\left(\frac{1}{2}-\frac{2}{3}\right)+\left(\frac{2}{3}-\frac{3}{4}\right)+\left(\frac{3}{4}-\frac{4}{5}\right)+\ldots =\frac{1}{2} , (Cancel the repeated terms with opposite signs and the last term that is not canceled converges to 0 0 )

We can have ∑ n = 1 ∞ 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) = 1 3 × ∑ n = 1 ∞ [ 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) − 1 ( n + 2 ) ( n + 3 ) ( n + 4 ) ] \sum_{n=1}^\infty \frac{1}{(n+1)(n+2)(n+3)(n+4)}\\ =\frac{1}{3}\times\sum_{n=1}^\infty \left[\frac{1}{(n+1)(n+2)(n+3)}-\frac{1}{(n+2)(n+3)(n+4)}\right]\\ Using similar strategy, the expression equals to 1 3 × 1 2 ⋅ 3 ⋅ 4 = 1 72 \frac{1}{3}\times\frac{1}{2 \cdot 3 \cdot 4}=\frac{1}{72} , it turns out that a + b = 73 a+b=73 .

19 Helpful 0 Interesting 0 Brilliant 0 Confused

Ah, why didn't I think of that earlier?! This makes the problem a one-liner and is much more elegant than my approach. Upvoted. :)

And just for the record, this "strategy" is called a telescoping sum where all the subsequent terms of the sum gets cancelled out.

Prasun Biswas - 6 years, 3 months ago

Nice Solution :D

Paul Ryan Longhas - 6 years, 3 months ago

Nice solution,anyway,can you teach me about calculus?

Frankie Fook - 6 years, 3 months ago
Bhargav Upadhyay
Feb 27, 2015

∑ n = 1 ∞ 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) = 1 6 ( n + 1 ) + − 1 2 ( n + 2 ) + 1 2 ( n + 3 ) + − 1 6 ( n + 4 ) = ∑ n = 1 ∞ 1 6 ( 1 ( n + 1 ) − 1 ( n + 4 ) ) − ∑ n = 1 ∞ 1 2 ( 1 ( n + 2 ) − 1 ( n + 3 ) ) = 1 6 ( 1 2 + 1 3 + 1 4 ) − 1 2 ( 1 3 ) = 1 6 ( 13 12 − 1 ) = 1 72 ∴ a = 1 , b = 72 ∴ a + b = 73 \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (n+1)(n+2)(n+3)(n+4) } } \\ \frac { 1 }{ (n+1)(n+2)(n+3)(n+4) } \quad =\quad \frac { 1 }{ 6(n+1) } +\frac { -1 }{ 2(n+2) } +\frac { 1 }{ 2(n+3) } +\frac { -1 }{ 6(n+4) } \\ =\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ 6 } } (\frac { 1 }{ (n+1) } -\frac { 1 }{ (n+4) } )\quad -\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ 2 } } (\frac { 1 }{ (n+2) } -\frac { 1 }{ (n+3) } )\\ =\frac { 1 }{ 6 } (\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 4 } )\quad -\quad \frac { 1 }{ 2 } (\frac { 1 }{ 3 } )\quad =\quad \frac { 1 }{ 6 } (\frac { 13 }{ 12 } -1)\quad =\quad \frac { 1 }{ 72 } \\ \therefore a=1,\quad b=72\\ \therefore a+b\quad =\quad 73\\

7 Helpful 0 Interesting 1 Brilliant 0 Confused

Since I'm lazy to use partial fractional decomposition method, I used a shorter approach. Notice that ( n + 1 ) ( n + 4 ) (n+1)(n+4) and ( n + 2 ) ( n + 3 ) (n+2)(n+3) when expanded have every term same except the constant term at the end. The constants differ by 2 2 . We use this fact and first break the general term as follows:

f ( n ) = 1 ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) ⟹ f ( n ) = 1 2 ⋅ ( 1 ( n + 1 ) ( n + 4 ) − 1 ( n + 2 ) ( n + 3 ) ) f(n)=\frac{1}{(n+1)(n+2)(n+3)(n+4)}\\ \implies f(n)=\frac{1}{2}\cdot \left(\frac{1}{(n+1)(n+4)}-\frac{1}{(n+2)(n+3)}\right)

Now, we can simply break each of the two fractions inside f ( n ) f(n) by introducing special zero values in the numerator and the whole general term breaks as:

f ( n ) = 1 2 { 1 3 ( 1 n + 1 − 1 n + 4 ) − ( 1 n + 2 − 1 n + 3 ) } f(n)=\frac{1}{2}\left\{ \frac{1}{3}\left(\frac{1}{n+1}-\frac{1}{n+4}\right)-\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\right\}

Rearranging the terms gives us the same as you evaluated using partial fractions. Although I think that rearranging it is redundant and one can calculate the sum just by telescoping each of the smaller sums of the general term.

Prasun Biswas - 6 years, 3 months ago

Hey really good observation... I find it very quick and easy to do partial fraction of first order polynomial so gone for it..

Bhargav Upadhyay - 6 years, 3 months ago

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