You Gold to Be Kidding Me

Number Theory Level 5

Let ϕ ( N ) \phi \ (N) denote the number of positive integers less than N N and prime to it. Then evaluate the value of:-

ϕ ( 1 ) 5 5 2 + 1 − ϕ ( 3 ) 125 12 5 2 + 1 + ϕ ( 5 ) 3125 312 5 2 + 1 + … \phi \ (1) \ \frac{5}{5^2 + 1} \ - \phi \ (3) \ \frac{125}{125^2+1} \ + \phi \ (5) \ \frac{3125}{3125^2+1} + \ldots

Note :- You might want to use a calculator in the last step. Round your answer to 3 3 decimal places.


The answer is 0.177514.

Kunal Verma by Kunal Verma , 22, India

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1 solution

Advait Nene
May 9, 2020

We can rewrite the sum as follows: ∑ n = 0 ∞ ( − 1 ) n ϕ ( 2 n + 1 ) 5 2 n + 1 5 2 ( 2 n + 1 ) + 1 \sum_{n=0}^{\infty}(-1)^{n}\phi(2n+1)\frac{5^{2n+1}}{5^{2(2n+1)}+1}

Consider x x 2 + 1 \frac{x}{x^{2}+1} . Dividing the numerator and denominator by x 2 x^{2} , we get x x 2 + 1 = 1 x 1 x 2 + 1 \frac{x}{x^{2}+1}=\frac{\frac{1}{x}}{\frac{1}{x^{2}}+1} We can rewrite this as the sum of the geometric series ∑ k = 0 ∞ ( − 1 ) k ( 1 x ) 2 k + 1 \sum_{k=0}^{\infty}(-1)^{k}\left(\frac{1}{x}\right)^{2k+1}

Therefore, we can rewrite the expression 5 2 n + 1 5 2 ( 2 n + 1 ) + 1 = ∑ k = 0 ∞ ( − 1 ) k ( 1 5 2 n + 1 ) 2 k + 1 = ∑ k = 0 ∞ ( − 1 ) k ( 1 5 ) ( 2 n + 1 ) ( 2 k + 1 ) \frac{5^{2n+1}}{5^{2(2n+1)}+1}=\sum_{k=0}^{\infty}(-1)^{k}\left(\frac{1}{5^{2n+1}}\right)^{2k+1}=\sum_{k=0}^{\infty}(-1)^{k}\left(\frac{1}{5}\right)^{(2n+1)(2k+1)}

Therefore, we can write the whole expression as ∑ n = 0 ∞ ( − 1 ) n ϕ ( 2 n + 1 ) 5 2 n + 1 5 2 ( 2 n + 1 ) + 1 = ∑ n = 0 ∞ ( − 1 ) n ϕ ( 2 n + 1 ) ∑ k = 0 ∞ ( − 1 ) k ( 1 5 ) ( 2 n + 1 ) ( 2 k + 1 ) = ∑ n = 0 ∞ ϕ ( 2 n + 1 ) ∑ k = 0 ∞ ( − 1 ) n + k ( 1 5 ) ( 2 n + 1 ) ( 2 k + 1 ) \sum_{n=0}^{\infty}(-1)^{n}\phi(2n+1)\frac{5^{2n+1}}{5^{2(2n+1)}+1}=\sum_{n=0}^{\infty}(-1)^{n}\phi(2n+1)\sum_{k=0}^{\infty}(-1)^{k}\left(\frac{1}{5}\right)^{(2n+1)(2k+1)}=\sum_{n=0}^{\infty}\phi(2n+1)\sum_{k=0}^{\infty}(-1)^{n+k}\left(\frac{1}{5}\right)^{(2n+1)(2k+1)}

Writing out the first few terms, ϕ ( 1 ) ( ( 1 5 ) 1 − ( 1 5 ) 3 + ( 1 5 ) 5 − . . . ) + ϕ ( 3 ) ( − ( 1 5 ) 3 + ( 1 5 ) 9 − ( 1 5 ) 15 + . . . ) + ϕ ( 5 ) ( ( 1 5 ) 5 − ( 1 5 ) 15 + ( 1 5 ) 25 − . . . ) \phi(1)\left(\left(\frac{1}{5}\right)^{1}-\left(\frac{1}{5}\right)^{3}+\left(\frac{1}{5}\right)^{5}-...\right)+\phi(3)\left(-\left(\frac{1}{5}\right)^{3}+\left(\frac{1}{5}\right)^{9}-\left(\frac{1}{5}\right)^{15}+...\right)+\phi(5)\left(\left(\frac{1}{5}\right)^{5}-\left(\frac{1}{5}\right)^{15}+\left(\frac{1}{5}\right)^{25}-...\right)

Regrouping, we see that our sum is equal to ( 1 5 ) 1 ϕ ( 1 ) − ( 1 5 ) 3 ( ϕ ( 1 ) + ϕ ( 3 ) ) + ( 1 5 ) 5 ( ϕ ( 1 ) + ϕ ( 5 ) ) − ( 1 5 ) 7 ( ϕ ( 1 ) + ϕ ( 7 ) ) + ( 1 5 ) 9 ( ϕ ( 1 ) + ϕ ( 3 ) + ϕ ( 9 ) ) − . . . \left(\frac{1}{5}\right)^{1}\phi(1)-\left(\frac{1}{5}\right)^{3}(\phi(1)+\phi(3))+\left(\frac{1}{5}\right)^{5}(\phi(1)+\phi(5))-\left(\frac{1}{5}\right)^{7}(\phi(1)+\phi(7))+\left(\frac{1}{5}\right)^{9}(\phi(1)+\phi(3)+\phi(9))-...

In general, we get that our sum is equal to ∑ n = 0 ∞ ( − 1 ) n ( 1 5 ) 2 n + 1 ∑ d ∣ 2 n + 1 ϕ ( d ) = ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ( 1 5 ) 2 n + 1 \sum_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{5}\right)^{2n+1}\sum_{d|2n+1}\phi(d)=\sum_{n=0}^{\infty}(-1)^{n}(2n+1)\left(\frac{1}{5}\right)^{2n+1} , using the identity ∑ d ∣ n ϕ ( d ) = n \sum_{d|n}\phi(d)=n . Let's set x = 1 5 x=\frac{1}{5} and consider the sum ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) x 2 n + 1 \sum_{n=0}^{\infty}(-1)^{n}(2n+1)x^{2n+1}

∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) x 2 n + 1 = x ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) x 2 n = x d d x ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 = x d d x x x 2 + 1 = x − x 3 ( x 2 + 1 ) 2 \sum_{n=0}^{\infty}(-1)^{n}(2n+1)x^{2n+1}=x\sum_{n=0}^{\infty}(-1)^{n}(2n+1)x^{2n}=x\frac{d}{dx}\sum_{n=0}^{\infty}(-1)^{n}x^{2n+1}=x\frac{d}{dx}\frac{x}{x^{2}+1}=\frac{x-x^{3}}{(x^{2}+1)^{2}}

Plugging in x = 1 5 x=\frac{1}{5} , we get that our sum is equal to 1 5 − 1 125 2 6 2 2 5 2 = 24 125 2 6 2 2 5 2 = 24 ⋅ 5 2 6 2 = 30 169 = 0.177514 \frac{\frac{1}{5}-\frac{1}{125}}{\frac{26^{2}}{25^{2}}}=\frac{\frac{24}{125}}{\frac{26^{2}}{25^{2}}}=\frac{24\cdot5}{26^{2}}=\boxed{\frac{30}{169}=0.177514}

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