"Prove ∑k=1∞(−1)kk∑j=12k(−1)jj=π248+14ln22" solved and explained

Zeenat Horn

Answered question

2022-02-24

Prove

\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k} \sum_{j = 1}^{2 k} \frac{{(- 1)}^{j}}{j} = \frac{π^{2}}{48} + \frac{1}{4} \ln^{2} 2

Kathryn Duggan

Beginner2022-02-25Added 7 answers

Use

\sum_{j = 1}^{n} \frac{{(- 1)}^{j - 1}}{j} = \ln 2 + {(- 1)}^{n - 1} \int_{0}^{1} \frac{x^{n}}{1 + x} d x

we get

\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k} \sum_{j = 1}^{2 k} \frac{{(- 1)}^{j}}{j} = \ln 2 \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k} - \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k} \int_{0}^{1} \frac{x^{2 k}}{1 + x} d x

= \ln^{2} 2 + \int_{0}^{1} \frac{1}{1 + x} \sum_{k = 1}^{\infty} \frac{{(- x^{2})}^{k}}{k} d x = \ln^{2} 2 - \int_{0}^{1} \frac{\ln (1 + x^{2})}{1 + x} d x

let

f (t) = \int_{0}^{1} \frac{\ln (1 + t x^{2})}{1 + x} d x

then

f^{'} (t) = \int_{0}^{1} \frac{x^{2}}{(1 + x) (1 + t x^{2})} d x

= \frac{1}{t + 1} \int_{0}^{1} \frac{x - 1}{1 + t x^{2}} d x + \frac{1}{t + 1} \int_{0}^{1} \frac{d x}{x + 1}

= \frac{1}{t + 1} {[\frac{1}{2 t} \ln (1 + t x^{2}) - \frac{1}{\sqrt{t}} \arctan (\sqrt{t} x) + \ln (x + 1)]}_{0}^{1}

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Tye Rhodes

Beginner2022-02-26Added 4 answers

Note that

\sum_{j = 1}^{2 k} \frac{{(- 1)}^{j}}{j} = \sum_{j = 1}^{k} \frac{1}{2 j} - \sum_{j = 1}^{k} \frac{1}{2 j - 1} = \frac{1}{2} \sum_{j = 1}^{k} \frac{1}{j} - (\sum_{j = 1}^{2 k} \frac{1}{j} - \sum_{j = 1}^{k} \frac{1}{2 j}) = \sum_{j = 1}^{k} \frac{1}{j} - \frac{j = 1}{2 k} \frac{1}{j}

= H_{k} - H_{2 k}

Then,

\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k} \sum_{j = 1}^{2 k} \frac{{(- 1)}^{j}}{j} = \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k} (H_{k} - H_{2 k})

= \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k} H_{k} - 2 \sum_{k = 1}^{\infty} \frac{i^{2 k}}{2 k} H_{2 k}

= f (- 1) - 2 R (f (i))

where

f (x) = \sum_{k = 1}^{\infty} \frac{x^{k}}{k} H_{k}

However,

f (x) = \sum_{k = 1}^{\infty} x^{k} H_{k} \int_{0}^{1} t^{k - 1} d t = \int_{0}^{1} \frac{1}{t} \sum_{k = 1}^{\infty} {(x t)}^{k} H_{k} d t = \int_{0}^{1} \frac{1}{t} [- \frac{\ln (1 - x t)}{1 - x t}] d t

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