Calculus Problem by LOGA 6

Calculus Level 4

Let a n = ∑ k = 1 n 1 k \displaystyle a_n=\sum_{k=1}^{n}\frac{1}{k} and f ( x ) = ∑ n = 1 ∞ a n x n \displaystyle f(x)=\sum_{n=1}^{\infty}a_nx^n . f ( 1 2 ) + f ( 3 4 ) + f ( 7 8 ) = c ln ⁡ 2 f\left(\frac{1}{2}\right)+f\left(\frac{3}{4}\right)+f\left(\frac{7}{8}\right)=c\ln{2} where c c is a positive integer. Find c c .


The answer is 34.

Logic Alpha by Logic Alpha , 19, USA

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1 solution

Mark Hennings
Nov 21, 2020

Note that ( 1 − x ) f ( x ) = ∑ n = 1 ∞ a n x n − ∑ n = 1 ∞ a n x n + 1 = x + ∑ n = 2 ∞ ( a n − a n − 1 ) x n = ∑ n = 1 ∞ x n n = − ln ⁡ ( 1 − x ) (1-x)f(x) \; = \; \sum_{n=1}^\infty a_n x^n - \sum_{n=1}^\infty a_nx^{n+1} \; =\; x + \sum_{n=2}^\infty(a_n - a_{n-1})x^n \; = \; \sum_{n=1}^\infty \frac{x^n}{n} \; = \; -\ln(1-x) for − 1 < x < 1 -1 < x < 1 , and hence f ( x ) = − ln ⁡ ( 1 − x ) 1 − x − 1 < x < 1 f(x) \; = \; -\frac{\ln(1-x)}{1-x} \hspace{2cm} -1 < x < 1 Hence f ( 1 2 ) + f ( 3 4 ) + f ( 7 8 ) = 34 ln ⁡ 2 f(\tfrac12) + f(\tfrac34) + f(\tfrac78) = \boxed{34}\ln2 .

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