Strange Infinite Series

Calculus Level 4

\begin{aligned} \sum_{\substack{n>1\\ m>1}}^{\infty} \frac{1}{n^{m}-1} &=\frac{1}{2^2-1} + \frac{1}{2^3-1} + \frac{1}{3^2-1} + \frac{1}{4^2-1} + \cdots \\ &=\frac{1}{3} + \frac{1}{7} + \frac{1}{8} + \frac{1}{15} + \cdots \\\\ &=\, ? \end{aligned}

Note:

  • In the above infinite series, n n and m m are positive integers greater than 1.
  • Double-counting is not allowed, i.e. if n m = m n , n^m=m^n, only one of 1 n m − 1 \frac{1}{n^m-1} and 1 m n − 1 \frac{1}{m^n-1} appears in the series. For example, since 4 2 = 2 4 , 4^2=2^4, you would only count 1 16 − 1 = 1 15 \frac{1}{16-1}=\frac{1}{15} once.


The answer is 1.000.

Joshua Lowrance by Joshua Lowrance , 18, USA

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2 solutions

Joshua Lowrance
Aug 18, 2018

First, let's prove that ( 1 n + 1 n 2 + 1 n 3 \frac{1}{n} + \frac{1}{n^{2}} + \frac{1}{n^{3}} + ... ) = 1 n − 1 \frac{1}{n-1}

( 1 n + 1 n 2 + 1 n 3 \frac{1}{n} + \frac{1}{n^{2}} + \frac{1}{n^{3}} + ... ) = S 1 S_{1}

( 1 n 2 + 1 n 3 + 1 n 4 \frac{1}{n^{2}} + \frac{1}{n^{3}} + \frac{1}{n^{4}} + ... ) = S 1 − 1 n S_{1} - \frac{1}{n}

n ( 1 n 2 + 1 n 3 + 1 n 4 + . . . ) n(\frac{1}{n^{2}} + \frac{1}{n^{3}} + \frac{1}{n^{4}} + ... ) = n ( S 1 − 1 n ) n(S_{1} - \frac{1}{n})

( 1 n + 1 n 2 + 1 n 3 \frac{1}{n} + \frac{1}{n^{2}} + \frac{1}{n^{3}} + ... ) = n ( S 1 ) − 1 n(S_{1}) - 1

S 1 S_{1} = n ( S 1 ) − 1 n(S_{1}) - 1

1 1 = n ( S 1 ) − S 1 n(S_{1}) - S_{1}

1 1 = ( n − 1 ) ( S 1 ) (n-1)(S_{1})

1 n − 1 \frac{1}{n-1} = S 1 S_{1}

We can now use this to help solve the problem.

1 1 + 1 2 + 1 3 + 1 4 + 1 5 \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... = S 2 S_{2}

1 1 \frac{1}{1} +( 1 2 \frac{1}{2} + 1 4 \frac{1}{4} + 1 8 \frac{1}{8} + ... ) + ( 1 3 \frac{1}{3} + 1 9 \frac{1}{9} + 1 27 \frac{1}{27} + ... ) + ( 1 5 \frac{1}{5} + 1 25 \frac{1}{25} + 1 125 \frac{1}{125} + ... ) + ... = S 2 S_{2}

(We would leave out every ( 1 n m \frac{1}{n^{m}} + 1 ( n m ) 2 \frac{1}{(n^{m})^{2}} + 1 ( n m ) 3 \frac{1}{(n^{m})^{3}} + ... ) because we already covered all those terms in ( 1 n \frac{1}{n} + 1 n 2 \frac{1}{n^{2}} + 1 n 3 \frac{1}{n^{3}} + ... ). For instance, we would leave out ( 1 4 \frac{1}{4} + 1 16 \frac{1}{16} + 1 64 \frac{1}{64} + ... ) because ( 1 2 \frac{1}{2} + 1 4 \frac{1}{4} + 1 8 \frac{1}{8} + ... ) already covers all those terms, and we don't want to repeat terms. This way, we cover every term once, and only once.)

We can also write the previous step as this:

1 1 \frac{1}{1} +( 1 2 \frac{1}{2} + 1 4 \frac{1}{4} + 1 8 \frac{1}{8} + ... ) + ( 1 3 \frac{1}{3} + 1 9 \frac{1}{9} + 1 27 \frac{1}{27} + ... ) + ( 1 4 \frac{1}{4} + 1 16 \frac{1}{16} + 1 64 \frac{1}{64} + ... ) + ( 1 5 \frac{1}{5} + 1 25 \frac{1}{25} + 1 125 \frac{1}{125} + ... ) + ... - ( 1 4 \frac{1}{4} + 1 16 \frac{1}{16} + 1 64 \frac{1}{64} + ... ) - ( 1 8 \frac{1}{8} + 1 64 \frac{1}{64} + 1 256 \frac{1}{256} + ... ) - ( 1 9 \frac{1}{9} + 1 81 \frac{1}{81} + 1 729 \frac{1}{729} + ... ) - ( 1 16 \frac{1}{16} + 1 256 \frac{1}{256} + 1 4096 \frac{1}{4096} + ... ) - ... = S 2 S_{2}

1 1 \frac{1}{1} +( 1 1 \frac{1}{1} ) + ( 1 2 \frac{1}{2} ) + ( 1 3 \frac{1}{3} ) + ( 1 4 \frac{1}{4} ) + ... - ( 1 3 \frac{1}{3} ) - ( 1 7 \frac{1}{7} ) - ( 1 8 \frac{1}{8} ) - ( 1 15 \frac{1}{15} ) - ... = S 2 S_{2}

1 1 \frac{1}{1} + S 2 S_{2} - ( 1 3 \frac{1}{3} ) - ( 1 7 \frac{1}{7} ) - ( 1 8 \frac{1}{8} ) - ( 1 15 \frac{1}{15} ) - ... = S 2 S_{2}

1 1 \frac{1}{1} - ( 1 3 \frac{1}{3} ) - ( 1 7 \frac{1}{7} ) - ( 1 8 \frac{1}{8} ) - ( 1 15 \frac{1}{15} ) - ... = 0 0

1 1 = ( 1 3 \frac{1}{3} ) + ( 1 7 \frac{1}{7} ) + ( 1 8 \frac{1}{8} ) + ( 1 15 \frac{1}{15} ) + ...

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Rohan Shinde
Dec 30, 2018

I think you are talking about Goldbach-Euler Theorem.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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