Logarithms

Calculus Level 3

S n = ∑ i = 1 n 1 4 i 2 − 2 i \large S_{n}=\sum_{i=1}^{n}\frac{1}{4i^{2}-2i}

For S n S_n as defined above, find ⌊ 100 lim ⁡ n → ∞ S n ⌋ \displaystyle \left \lfloor 100 \lim_{n \to \infty}S_n \right \rfloor .

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


The answer is 69.

Maximos Stratis by maximos stratis , 21, Greece

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1 solution

Relevant wiki: Maclaurin Series

S n = ∑ i = 1 n 1 4 i 2 − 2 i = ∑ i = 1 n 1 2 i ( 2 i − 1 ) = ∑ i = 1 n ( 1 2 i − 1 − 1 2 i ) = 1 − 1 2 + 1 3 − 1 4 + ⋯ + 1 2 n − 1 − 1 2 n \begin{aligned} S_n & = \sum_{i=1}^n \frac 1{4i^2-2i} \\ & = \sum_{i=1}^n \frac 1{2i(2i-1)} \\ & = \sum_{i=1}^n \left( \frac 1{2i-1} - \frac 1{2i} \right) \\ & = 1 - \frac 12 + \frac 13 - \frac 14 + \cdots + \frac 1{2n-1} - \frac 1{2n} \end{aligned}

lim ⁡ n → ∞ S n = 1 − 1 2 + 1 3 − 1 4 + ⋯ By Maclaurin series: ln ⁡ ( 1 + x ) = ∑ k = 1 ∞ ( − 1 ) k + 1 x k k ; putting x = 1 = ln ⁡ 2 \begin{aligned} \lim_{n \to \infty} S_n & = 1 - \frac 12 + \frac 13 - \frac 14 + \cdots & \small \color{#3D99F6} \text{By Maclaurin series: } \ln(1+x) = \sum_{k=1}^\infty \frac {(-1)^{k+1}x^k}k \text{; putting }x=1 \\ & = \ln 2 \end{aligned}

⟹ ⌊ 100 lim ⁡ n → ∞ S n ⌋ = ⌊ 100 ln ⁡ 2 ⌋ = 69 \implies \displaystyle \left \lfloor 100 \lim_{n \to \infty} S_n \right \rfloor = \left \lfloor 100 \ln 2 \right \rfloor = \boxed{69}

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