Find the sum

Calculus Level 3

S = 1 1 ⋅ 2 + 2 2 ⋅ 3 + 3 3 ⋅ 4 + 4 4 ⋅ 5 + 0 5 ⋅ 6 + 1 6 ⋅ 7 + . . . = ∑ k = 1 ∞ k mod 5 k ( k + 1 ) \begin{aligned} S&=\frac{1}{1\cdot2}+\frac{2}{2\cdot3}+\frac{3}{3\cdot4}+\frac{4}{4\cdot5}+\frac{0}{5\cdot6}+\frac{1}{6\cdot7}+...\\ &=\sum_{k=1}^∞\frac{k\text { mod 5}}{k(k+1)}\end{aligned}

Evaluate the infinite sum above and enter ⌊ 1 0 10 S ⌋ \lfloor 10^{10} S \rfloor .

Notation : ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function.


The answer is 16094379124.

Daniel Juncos by Daniel Juncos , 45, USA

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1 solution

Chew-Seong Cheong
Apr 22, 2017

S = lim ⁡ n → ∞ ( 1 1 ⋅ 2 + 2 2 ⋅ 3 + 3 3 ⋅ 4 + 4 4 ⋅ 5 + 0 5 ⋅ 6 + 1 6 ⋅ 7 + . . . + 4 ( 5 n − 1 ) ⋅ 5 n + 0 5 n ⋅ ( 5 n + 1 ) ) = lim ⁡ n → ∞ ( 1 1 − 1 2 + 2 2 − 2 3 + 3 3 − 3 4 + 4 4 − 4 5 + 0 + 1 6 − 1 7 + . . . + 4 5 n − 1 − 4 5 n + 0 ) = lim ⁡ n → ∞ ( 1 1 + 1 2 + 1 3 + 1 4 + 1 5 − 5 5 + 1 6 + 1 7 + . . . + 4 5 n − 1 + 1 5 n − 5 5 n ) = lim ⁡ n → ∞ [ ( 1 1 + 1 2 + 1 3 + . . . + 1 5 n ) − 5 ( 1 5 + 1 10 + 1 15 + . . . + 1 5 n ) ] = lim ⁡ n → ∞ [ ( 1 1 + 1 2 + 1 3 + . . . + 1 5 n ) − ( 1 1 + 1 2 + 1 3 + . . . + 1 n ) ] = lim ⁡ n → ∞ ( H 5 n − H n ) See note below. = ln ⁡ ( 5 n ) + γ − ln ⁡ ( n ) − γ = ln ⁡ 5 \begin{aligned} S &= \lim_{n \to \infty} \left( \frac 1{1\cdot2} + \frac 2{2\cdot3} +\frac 3{3\cdot4} + \frac 4{4\cdot5} +\frac 0{5\cdot6} + \frac 1{6\cdot7} +... + \frac 4{(5n-1)\cdot5n} + \frac 0{5n\cdot(5n+1)} \right) \\ &= \lim_{n \to \infty} \left( \frac 11 - \frac 12 +\frac 22 - \frac 23 +\frac 33- \frac 34 +\frac 44 - {\color{#3D99F6}\frac 45+ 0} + \frac 16 - \frac 17 +... + \frac 4{5n-1} - {\color{#3D99F6}\frac 4{5n}+ 0} \right) \\ &= \lim_{n \to \infty} \left(\frac 11 +\frac 12 +\frac 13+\frac 14 {\color{#3D99F6}+ \frac 15 - \frac 55} + \frac 16 + \frac 17 +... + \frac 4{5n-1} {\color{#3D99F6}+ \frac 1{5n} - \frac 5{5n}} \right) \\ &= \lim_{n \to \infty} \left[\left(\frac 11 +\frac 12 +\frac 13+...+ \frac 1{5n} \right) - 5 \left(\frac 15 +\frac 1{10} +\frac 1{15}+... + \frac 1{5n} \right) \right] \\ &= \lim_{n \to \infty} \left[ \left(\frac 11 +\frac 12 +\frac 13+...+ \frac 1{5n} \right) - \left(\frac 11 +\frac 12 +\frac 13+... + \frac 1n \right) \right] \\ &= \lim_{n\to \infty} (H_{5n} - H_n) & \small \color{#3D99F6} \text{See note below.} \\ &= \ln(5n) + \gamma - \ln(n)-\gamma \\ &= \ln 5 \end{aligned}

Note : lim ⁡ n → ∞ H n = ln ⁡ ( n ) − γ \displaystyle \lim_{n \to \infty} H_n = \ln (n) - \gamma , where H n H_n is the n n th harmonic number and γ \gamma is the Euler-Mascheroni constant .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

( 1 1 + 1 2 + 1 3 + . . . ) − 5 ( 1 5 + 1 10 + 1 15 + . . , ) = ∞ − ∞ \left(\frac 11 +\frac 12 +\frac 13+... \right) - 5 \left(\frac 15 +\frac 1{10} +\frac 1{15}+..,\right) = \infty - \infty

Pi Han Goh - 4 years, 1 month ago

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I have done the changes.

Chew-Seong Cheong - 4 years, 1 month ago

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lim ⁡ n → ∞ ( 1 1 + 1 2 + 1 3 + . . . + 1 5 n ) − lim ⁡ n → ∞ 5 ( 1 5 + 1 10 + 1 15 + . . . + 1 5 n ) = ∞ − ∞ \lim_{n \to \infty} \left(\frac 11 +\frac 12 +\frac 13+...+ \frac 1{5n} \right) - \lim_{n \to \infty} 5 \left(\frac 15 +\frac 1{10} +\frac 1{15}+... + \frac 1{5n} \right) = \infty - \infty

Pi Han Goh - 4 years, 1 month ago

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@Pi Han Goh – The mechanics aren't rigorous ( there shouldn't be an "n" remaining after a limit is taken, and it doesn't make sense to say that lim ⁡ n → ∞ H n = l n ( n ) − γ \lim_{n\to\infty} H_n=ln(n)-\gamma ) but his general idea is correct (and quite creative!).

If we call S n = ∑ k = 1 n k mod 5 k ⋅ ( k + 1 ) S_n = \sum_{k=1}^{n}{\frac{k\text{ mod 5}}{k\cdot{(k+1)}}} , then through Chew-Seong Cheong's methods from the first three lines, we can see that the following holds:

S 5 = 1 1 + 1 2 + 1 3 + 1 4 + 1 5 − 5 5 S_5 = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{5}{5}

S 10 = 1 1 + 1 2 + 1 3 + 1 4 + 1 5 − 5 5 + 1 6 + 1 7 + 1 8 + 1 9 + 1 10 − 5 10 S_{10} = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{5}{5}+ \frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}-\frac{5}{10}

= ( 1 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 + 1 9 + 1 10 ) − ( 1 1 + 1 2 ) =( \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+ \frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10})-(\frac{1}{1}+\frac{1}{2})

As long as the index is a multiple of 5, the above structure holds. So we guarantee that by defining T n = S 5 n T_n = S_{5n} ; i.e.

T n = ( 1 1 + 1 2 + 1 3 + . . . + 1 5 n ) − ( 1 1 + 1 2 + . . . + 1 n ) T_n =( \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+ ... + \frac{1}{5n}) - ( \frac{1}{1}+\frac{1}{2}+...+ \frac{1}{n})

So,

= [ 1 1 + 1 2 + 1 3 + . . . + 1 5 n − l n ( 5 n ) ] − [ 1 1 + 1 2 + . . . + 1 n − l n ( n ) ] + l n ( 5 n ) − l n ( n ) =[\: \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+ ... + \frac{1}{5n}-ln(5n)\:]-[\: \frac{1}{1}+\frac{1}{2}+ ... + \frac{1}{n}-ln(n)\:] +ln(5n)-ln(n)

= [ 1 1 + 1 2 + 1 3 + . . . + 1 5 n − l n ( 5 n ) ] − [ 1 1 + 1 2 + . . . + 1 n − l n ( n ) ] + l n ( 5 ) =[\: \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+ ... + \frac{1}{5n}-ln(5n)\:]-[\: \frac{1}{1}+\frac{1}{2}+ ... + \frac{1}{n}-ln(n)\:] +ln(5)

As n → ∞ {n\to\infty} , then 5 n → ∞ {5n\to\infty} , so the first portion of T n T_n goes to γ \gamma , as does the second portion. Since both exist, we can subtract one limit from the other:

lim ⁡ n → ∞ S n = lim ⁡ n → ∞ T n = γ − γ + l n ( 5 ) = l n ( 5 ) \lim_{n\to\infty} S_n=\lim_{n\to\infty} T_n = \gamma - \gamma + ln(5) = ln(5)

Daniel Juncos - 4 years, 1 month ago

Ah, I follow it now. Excellent. My proof is completely different.

Daniel Juncos - 4 years, 1 month ago

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