Best solutions to - "Findlimn→∞∑i=1n2n(n+2i)2"

palmantkf4u

Answered question

2022-04-05

Find

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

Answer & Explanation

Drahthaare89c

Beginner2022-04-06Added 19 answers

The expression

\sum_{i = 1}^{n} \frac{2 n}{{(n + 2 i)}^{2}} = \frac{1}{n} \sum_{i = 1}^{n} \frac{2}{{(1 + \frac{2 i}{n})}^{2}}

is the Riemann sum of the function

f (x) = \frac{2}{{(1 + 2 x)}^{2}}

over the interval

I = [0, 1]

, corresponding to the uniform partition of I into n equal parts. Since f is Riemann-integrable (being a continuous function over a closed and bounded interval), this sum approaches the integral of f over I as

n \to \infty

. That is,

lim_{n \to \infty} \frac{1}{n} \sum_{i = 1}^{n} \frac{2}{{(1 + \frac{2 i}{n})}^{2}} = \int_{0}^{1} \frac{2}{{(1 + 2 x)}^{2}} d x

= - \frac{1}{1 + 2 x} ∣_{x = 0}^{x = 1} = \frac{2}{3}

Alannah Campos

Beginner2022-04-07Added 10 answers

You can also use Euler-Maclaurin summation.
The first-order Euler-Maclaurin formula says

\sum_{i = 1}^{n} f (i) = \int_{1}^{n} f (x), d x + {f (1) + f (n) 2 + \int_{1}^{n} f^{'} (x) (x - ⌊ x ⌋ - \frac{1}{2}), d x .

Since

| x - ⌊ x ⌋ - \frac{1}{2} | \leq 1

, with

f (x) = \frac{2 n}{{(n + 2 x)}^{2}}

we have

\sum_{i = 1}^{n} \frac{2 n}{{(n + 2 i)}^{2}} = \int_{1}^{n} \frac{2 n, d x}{{(n + 2 x)}^{2}} + R_{n},

where

| R_{n} | \leq | \frac{3 f (n) - f (1)}{2} | = | \frac{3}{n} - \frac{n}{{(n + 2)}^{2}} |

Therefore,

lim_{n \to \infty} \sum_{i = 1}^{n} \frac{2 n}{{(n + 2 i)}^{2}} = lim_{n \to \infty} \int_{1}^{n} \frac{2 n, d x}{{(n + 2 x)}^{2}}

= lim_{n \to \infty} {[\frac{- n}{n + 2 x}]}_{1}^{n}

= lim_{n \to \infty} (- \frac{1}{3} + \frac{n}{n + 2})

= \frac{2}{3}

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