Maybe a quadruple summation

Calculus Level 5

lim ⁡ i → ∞ ( ∑ j = 1 j = i ( ( ∑ k = 1 k = j 1 lim ⁡ l → ∞ ∑ m = 1 m = l 1 ∏ n = 0 n = k m + n ) + 1 ) ( ( j + 1 ) ! ) 2 ) \displaystyle \LARGE \lim _{ i\rightarrow \infty }{ \left( \displaystyle \sum _{ j=1 }^{ j=i }{ \frac { \left( \left( \sum _{ k=1 }^{ k=j }{ \frac { 1 }{ \lim _{ l\rightarrow \infty }{ \sum _{ m=1 }^{ m=l }{ \frac { 1 }{ \prod _{ n=0 }^{ n=k }{ m+n } } } } } } \right) +1 \right) }{ { \left( (j+1)! \right) }^{ 2 } } } \right) }

If the answer can be expressed as e a − b {e}^{a}-b , select a + b a+b .

Try its sister problem here
2 3 8 7 4 6 5 1

Abhishek Sharma by Abhishek Sharma , 22, India

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1 solution

Kartik Sharma
May 10, 2015

Nice easy problem!

Well, we'll start with the product at the bottom.

1 ∏ n = 0 k ( m + n ) = ( m − 1 ) ! ( m + k ) ! \displaystyle \frac{1}{\prod_{n=0}^{k}{(m+n)}} = \frac{(m-1)!}{(m+k)!} which is quite easy to observe.

Now the next summation,

lim ⁡ l → ∞ ∑ m = 1 l Γ ( m ) Γ ( m + k + 1 ) \displaystyle \lim_{l \rightarrow \infty}{\sum_{m=1}^{l}{\frac{\Gamma(m)}{\Gamma(m+k+1)}}}

Multiply and divide by Γ ( k + 1 ) \Gamma(k+1) and take out the denominator while let the numerator be inside the summation.

1 Γ ( k + 1 ) lim ⁡ l → ∞ ∑ m = 1 l Γ ( m ) Γ ( k + 1 ) Γ ( m + k + 1 ) \displaystyle \frac{1}{\Gamma(k+1)}\lim_{l \rightarrow \infty}{\sum_{m=1}^{l}{\frac{\Gamma(m)\Gamma(k+1)}{\Gamma(m+k+1)}}}

Then we can substitute Beta function inside summation,

1 Γ ( k + 1 ) lim ⁡ l → ∞ ∑ m = 1 l ∫ 0 1 t m − 1 ( 1 − t ) k \displaystyle \frac{1}{\Gamma(k+1)}\lim_{l \rightarrow \infty}{\sum_{m=1}^{l}{\int_{0}^{1}{{t}^{m-1}{(1-t)}^{k}}}}

∫ 0 1 ( 1 − t ) k ∑ m = 1 ∞ t m − 1 \displaystyle \int_{0}^{1}{{(1-t)}^{k}\sum_{m=1}^{\infty}{{t}^{m-1}}}

∫ 0 1 ( 1 − t ) k − 1 = 1 k \displaystyle \int_{0}^{1}{{(1-t)}^{k-1}} = \frac{1}{k}

Putting everything in the second summation,

∑ k = 1 j 1 1 k . k ! \displaystyle \sum_{k=1}^{j}{\frac{1}{\frac{1}{k.k!}}}

∑ k = 1 j k . k ! = ∑ k = 1 j ( k + 1 ) ! − k ! \displaystyle \sum_{k=1}^{j}{k.k!} = \sum_{k=1}^{j}{(k+1)! - k!}

= ( j + 1 ) ! − 1 \displaystyle = (j+1)! - 1

Finally into the last summation,

∑ j = 1 ∞ 1 ( j + 1 ) ! \displaystyle \sum_{j = 1}^{\infty}{\frac{1}{(j+1)!}}

e − 2 \displaystyle \boxed{e - 2}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

You could've done the first part like this.

lim ⁡ l → ∞ ∑ m = 1 m = l 1 ∏ n = 0 n = k m + n \lim _{ l\rightarrow \infty }{ \sum _{ m=1 }^{ m=l }{ \frac { 1 }{ \prod _{ n=0 }^{ n=k }{ m+n } } } } lim ⁡ l → ∞ ∑ m = 1 m = l 1 m × ( m + 1 ) × ( m + 2 ) … ( m + k ) \lim _{ l\rightarrow \infty }{ \sum _{ m=1 }^{ m=l }{ \frac { 1 }{ m\times \left( m+1 \right) \times \left( m+2 \right) \dots \left( m+k \right) } } } lim ⁡ l → ∞ ∑ m = 1 m = l 1 k ( m + k ) − m m × ( m + 1 ) × ( m + 2 ) … ( m + k ) \lim _{ l\rightarrow \infty }{ \sum _{ m=1 }^{ m=l }{ \frac { 1 }{ k } \frac { \left( m+k \right) -m }{ m\times \left( m+1 \right) \times \left( m+2 \right) \dots \left( m+k \right) } } } lim ⁡ l → ∞ 1 k ∑ m = 1 m = l 1 m × ( m + 1 ) … ( m + k − 1 ) − 1 ( m + 1 ) × … ( m + k ) \lim _{ l\rightarrow \infty }{ \frac { 1 }{ k } \sum _{ m=1 }^{ m=l }{ \frac { 1 }{ m\times \left( m+1 \right) \dots \left( m+k-1 \right) } -\frac { 1 }{ \left( m+1 \right) \times \dots \left( m+k \right) } } } lim ⁡ l → ∞ 1 k ( 1 1 × ( 2 ) × ( 3 ) … ( k ) − 1 ( l + 1 ) × ( l + 2 ) … ( l + k ) ) \lim _{ l\rightarrow \infty }{ \frac { 1 }{ k } \left( \frac { 1 }{ 1\times \left( 2 \right) \times \left( 3 \right) \dots \left( k \right) } -\frac { 1 }{ \left( l+1 \right) \times \left( l+2 \right) \dots \left( l+k \right) } \right) } 1 k ( k ! ) \frac { 1 }{ k\left( k! \right) }

Abhishek Sharma - 6 years, 1 month ago

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Oh yeah, it's just the same thing that Kishlaya's identity has 2 proofs - one relating to Beta function and the other(and more clever) using telescoping.

Kartik Sharma - 6 years, 1 month ago

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