Explained: "I need to find: limn→∞n∑k=n∞12k(2k+1)"

2alr8w

Answered question

2022-01-24

I need to find:

lim_{n \to \infty} n \sum_{k = n}^{\infty} \frac{1}{2 k (2 k + 1)}

Bottisiooq

Beginner2022-01-25Added 9 answers

Notice

\frac{1}{2} (\frac{1}{2 k - 1} - \frac{1}{2 k + 1})

\frac{1}{(2 k - 1) (2 k + 1)} \geq \frac{1}{2 k (2 k + 1)} \geq \frac{1}{2 k (2 k + 2)}

The partial sums start at

k = n

is squeezed between two telescoping series. This leads to

\frac{1}{4 n - 2} \geq \sum_{k = n}^{\infty} \frac{1}{2 k (2 k + 1)} \geq \frac{1}{4 n}

As a result,

| n \sum_{k = n}^{\infty} \frac{1}{2 k (2 k + 1)} - \frac{1}{4} | \leq \frac{1}{8 n - 4}

Since

lim_{n \to \infty} \frac{1}{8 n - 4} = 0

, we get

lim_{n \to \infty} n \sum_{k = n}^{\infty} \frac{1}{2 k (2 k + 1)} = \frac{1}{4}

search633504

Beginner2022-01-26Added 16 answers

This solution is based on the decomposition

\frac{1}{2 k (2 k + 1)} = \frac{1}{2 k} - \frac{1}{2 k + 1}

and on the expansion up to order

\frac{1}{n}

of the n-th harmonic number

H_{n}

as
Here we go:

\sum_{k = n}^{\infty} = \frac{1}{2 n} - \frac{1}{2 n + 1} + \frac{1}{2 n + 2} - \frac{1}{2 n + 3} + \dots

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

= \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{k} + \sum_{k = 1}^{2 n - 1} \frac{1}{k} = 2 \sum_{k = 1}^{n - 1} \frac{1}{2 k}

= - \log 2 + H_{2 n - 1} - H_{n - 1}

= - \log 2 + (γ + \log (2 n - 1) + \frac{1}{4 n - 2}) - (γ + \log (n - 1) + \frac{1}{2 n - 2}) + O (\frac{1}{n^{2}})

= - \log 2 + \log (\frac{2 n - 1}{n - 1}) + \frac{1}{4 n} - \frac{1}{2 n} + O (\frac{1}{n^{2}})

= - \log 2 + \log 2 + \log (1 + \frac{1}{2 (n - 1)}) - \frac{1}{4 n} + O (\frac{1}{n^{2}})

= \frac{1}{4 n} + O (\frac{1}{n^{2}})

Thus,

lim_{n \to \infty} n \sum_{k = n}^{\infty} \frac{1}{2 k (2 k + 1)} = \frac{1}{4}

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