1800 followers problem

Calculus Level 5

∑ k = 1 1800 ψ ( k ) \large \sum_{k=1}^{1800} \psi(k)

If the above sum can be expressed as a H b − c γ − d aH_b-c\gamma-d for some not necessarily distinct positive integers a , b , c a,b,c and d d , evaluate a + b + c + d a+b+c+d where b , c b,c is maximum possible.

Notations :

This problem is original.


The answer is 7200.

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Nihar Mahajan by Nihar Mahajan , 20, India

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3 solutions

Raghav Arora
Feb 19, 2016

ψ ( z + 1 ) = − γ + ∫ 0 1 1 − x 2 1 − x d x = H z − γ → ∑ k = 1 1800 ψ ( z ) = ∑ n = 1 1799 ( H n − γ ) = ∑ n = 0 1799 H n − 1800 γ = ( 1799 + 1 ) ( H 1799 + 1 ) − 1800 γ = 1800 H 1800 − 1800 γ − 1800 1800 + 1800 + 1800 + 1800 = 7200 \psi (z+1)\\ =-\gamma +\int _{ 0 }^{ 1 }{ \frac { { 1-x }^{ 2 } }{ 1-x } dx } \\ ={ H }_{ z }-\gamma \\ \rightarrow \sum _{ k=1 }^{ 1800 }{ \psi (z)\quad =\quad } \sum _{ n=1 }^{ 1799 }{ { (H }_{ n } } -\gamma )\\ =\sum _{ n=0 }^{ 1799 }{ { H }_{ n } } -\quad 1800\gamma \\ =(1799+1)({ H }_{ 1799+1 })\quad -1800\gamma \\ =1800{ H }_{ 1800 }-1800\gamma -1800\\ 1800+1800+1800+1800\\ =\boxed { 7200 }

8 Helpful 0 Interesting 0 Brilliant 0 Confused

It would be better if you provide proof for ψ ( z + 1 ) = H z − γ \psi(z+1)=H_z-\gamma . Anyways , a nice solution +1

Nihar Mahajan - 5 years, 3 months ago

ψ ( 1 + k ) = 1 k + ψ ( k ) \psi(1+k) = \dfrac{1}{k} + \psi(k)
ψ ( 1 ) = − γ \psi(1) = -\gamma
Using both these properties,
ψ ( 1 ) = − γ \psi(1) = -\gamma
ψ ( 2 ) = 1 − γ \psi(2) = 1 -\gamma
ψ ( 3 ) = 1 + 1 2 − γ \psi(3) = 1 + \dfrac{1}{2} - \gamma
. .
. .



ψ ( 1800 ) = 1 + 1 2 + … 1 1799 − γ \psi(1800) = 1 + \dfrac{1}{2} + \ldots \dfrac{1}{1799} - \gamma
Adding all the terms,
∑ k = 1 1800 ψ ( k ) = 1799 1 + 1798 2 + 1797 3 + … + 1 1799 − 1800 γ \displaystyle \sum_{k=1}^{1800}\psi(k) = \dfrac{1799}{1} + \dfrac{1798}{2} + \dfrac{1797}{3} + \ldots + \dfrac{1}{1799} - 1800\gamma
∑ k = 1 1800 ψ ( k ) = 1800 1 − 1 + 1800 2 − 1 + 1800 3 − 1 + … + 1800 1799 − 1 − 1800 γ \displaystyle \sum_{k=1}^{1800}\psi(k) = \dfrac{1800}{1} - 1 + \dfrac{1800}{2} - 1 + \dfrac{1800}{3} - 1 + \ldots + \dfrac{1800}{1799} - 1 -1800\gamma
∑ k = 1 1800 ψ ( k ) = 1800 ( 1 + 1 2 + 1 3 + … + 1 1799 ) − 1799 − 1800 γ \displaystyle \sum_{k=1}^{1800}\psi(k) = 1800\left(1 + \dfrac{1}{2} + \dfrac{1}{3} + \ldots + \dfrac{1}{1799}\right) - 1799 - 1800\gamma
∑ k = 1 1800 ψ ( k ) = 1800 H 1799 − 1799 − 1800 γ \displaystyle \sum_{k=1}^{1800}\psi(k) = 1800H_{1799} - 1799 - 1800\gamma
I reached this stage and input my solution as 7198 and wondered why my answer was wrong, annoyed Nihar( :P ) and got him to write the max condition.
∑ k = 1 1800 ψ ( k ) = 1800 H 1799 − 1799 − 1800 γ = 1800 H 1800 − 1800 − 1800 γ \displaystyle \sum_{k=1}^{1800}\psi(k) = 1800H_{1799} - 1799 - 1800\gamma = 1800H_{1800} - 1800 - 1800\gamma

In general,
∑ k = 1 n ψ ( k ) = n H n − 1 − ( n − 1 ) − n γ = n ( H n − γ − 1 ) \displaystyle \sum_{k=1}^{n}\psi(k) = nH_{n-1} - (n-1) - n\gamma = n(H_{n} - \gamma - 1)

The proof of the two properties I mentioned can be found on the Brilliant Digamma wiki.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Good usage of the properties of these functions.

Thank you for the solution and saving me from reports :) +1

Nihar Mahajan - 5 years, 3 months ago
Harsh Shrivastava
Feb 19, 2016

Claim : ψ ( k + 1 ) = ψ ( k ) + 1 k \psi(k+1) = \psi(k) + \dfrac{1}{k}

Proof : An easy proof can be found here.

Now using ψ ( 1 ) = − γ \psi(1) = - \gamma , we get ψ ( 2 ) = − γ + 1 = − γ + H 1 ⋯ ⋯ ( ∗ ) \psi(2) = -\gamma + 1 = -\gamma + H_{1} \cdots \cdots (*)

ψ ( 3 ) = ψ ( 2 ) + 1 / 2 = − γ + H 2 \psi(3) = \psi(2) + 1/2 = -\gamma + H_{2} [Using (*)]

ψ ( 4 ) = − γ + H 3 \psi(4) = -\gamma + H_{3}

ψ ( 5 ) = − γ + H 4 \psi(5) = -\gamma + H_{4}

Thus noticing the pattern, we get ψ ( n + 1 ) = − γ + H n \psi(n+1) = -\gamma + H_{n}

Thus ∑ k = 1 1800 ψ ( k ) = − 1800 γ + ∑ i = 1 1799 H i \displaystyle \sum_{k=1}^{1800} \psi(k) = -1800 \gamma + \sum_{i=1}^{1799} H_{i}

Claim : ∑ i = 1 n H i = ( n + 1 ) H n + 1 − ( n + 1 ) \sum_{i=1}^{n} H_{i} = (n+1)H_{n+1} - (n+1)

Proof : Induction will do the job. (Left as an exercise for the reader.)

Thus ∑ k = 1 1800 ψ ( k ) = − 1800 γ + ∑ i = 1 1799 H i = − 1800 γ + 1800 H 1800 − 1800 \displaystyle \sum_{k=1}^{1800} \psi(k) = -1800 \gamma + \sum_{i=1}^{1799} H_{i} = -1800 \gamma +1800H_{1800} -1800

Final answer = 1800 + 1800 + 1800 + 1800 = 1800 × 4 = 7200 1800+1800+1800+1800 = 1800 \times 4 = \boxed{\boxed{7200}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Great solution! +1

Nihar Mahajan - 5 years, 3 months ago

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Thanks for entertaining my questions.

asad bhai - 5 years, 3 months ago

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No problem! :)

Nihar Mahajan - 5 years, 3 months ago

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