How to prove the following sequence converges to 0.5? an=∫01nxn−11+xdx

We have: an=∫01nxn−11+xdx=xn1+x∣01+∫01xn(1+x)2dx=12+∫01xn(1+x)2dx Since ∫01xn(1+x)2dx≤∫01xndx=1n+1 it follows that limn∫01xn(1+x)2dx=0 Thus limnan=12

I apologize in advance, this is a lot of math and few words. ∫01nxn−11+xdx=∫01nxn−1∑k=0∞(−x)kdx =∑k=0∞∫01nxn−1+k(−1)kdx =∑k=0∞(−1)knn+k Now, your want limn→∞n∑k=0∞(−1)kn+k=limn→∞n(∑k=0∞1n+2k−1n+2k+1) =limn→∞n∑k=0∞1(n+2k)2+n+2k =limn→∞1n∑k=0∞1(n+2k)2 (1) =limn→∞1n∑k=0∞1(1+2kn)2 =∫0∞1(1+2x)2dx (2) =12∫0∞1(1+x)2dx =12(−1x+1)∣0∞ =12 (1) is obtained by realizing that n+2k is negligible in comparison to (n+2k)2 as n approaches ∞ (2) uses the well-known identity

Using integration by parts, we obtain ∫01nxn−11+xdx=xn1+x|01+∫01xn(1+x)2dx=12+rn where clearly 0&lt;rn≤∫01xndx=1n+1→0, as n→∞

We have: an=∫01nxn−11+xdx=xn1+x∣01+∫01xn(1+x)2dx=12+∫01xn(1+x)2dx Since ∫01xn(1+x)2dx≤∫01xndx=1n+1 it follows that limn∫01xn(1+x)2dx=0 Thus limnan=12

I apologize in advance, this is a lot of math and few words. ∫01nxn−11+xdx=∫01nxn−1∑k=0∞(−x)kdx =∑k=0∞∫01nxn−1+k(−1)kdx =∑k=0∞(−1)knn+k Now, your want limn→∞n∑k=0∞(−1)kn+k=limn→∞n(∑k=0∞1n+2k−1n+2k+1) =limn→∞n∑k=0∞1(n+2k)2+n+2k =limn→∞1n∑k=0∞1(n+2k)2 (1) =limn→∞1n∑k=0∞1(1+2kn)2 =∫0∞1(1+2x)2dx (2) =12∫0∞1(1+x)2dx =12(−1x+1)∣0∞ =12 (1) is obtained by realizing that n+2k is negligible in comparison to (n+2k)2 as n approaches ∞ (2) uses the well-known identity

Using integration by parts, we obtain ∫01nxn−11+xdx=xn1+x|01+∫01xn(1+x)2dx=12+rn where clearly 0&lt;rn≤∫01xndx=1n+1→0, as n→∞

Best solutions to - "How to prove the following sequence converges to..."

iocasq4

Answered question

2022-01-24

How to prove the following sequence converges to 0.5?

a_{n} = \int_{0}^{1} \frac{n x^{n - 1}}{1 + x} d x

Answer & Explanation

helsinka04

Beginner2022-01-25Added 11 answers

We have:

a_{n} = \int_{0}^{1} \frac{n x^{n - 1}}{1 + x} d x = \frac{x^{n}}{1 + x} ∣_{0}^{1} + \int_{0}^{1} \frac{x^{n}}{{(1 + x)}^{2}} d x = \frac{1}{2} + \int_{0}^{1} \frac{x^{n}}{{(1 + x)}^{2}} d x

Since

\int_{0}^{1} \frac{x^{n}}{{(1 + x)}^{2}} d x \leq \int_{0}^{1} x^{n} d x = \frac{1}{n + 1}

it follows that

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

Thus

lim_{n} a_{n} = \frac{1}{2}

Jaiden Conrad

Beginner2022-01-26Added 14 answers

I apologize in advance, this is a lot of math and few words.

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

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

= \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} n}{n + k}

Now, your want

lim_{n \to \infty} n \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{n + k} = lim_{n \to \infty} n (\sum_{k = 0}^{\infty} \frac{1}{n + 2 k} - \frac{1}{n + 2 k + 1})

= lim_{n \to \infty} n \sum_{k = 0}^{\infty} \frac{1}{{(n + 2 k)}^{2} + n + 2 k}

= lim_{n \to \infty} \frac{1}{n} \sum_{k = 0}^{\infty} \frac{1}{{(n + 2 k)}^{2}}

(1)

= lim_{n \to \infty} \frac{1}{n} \sum_{k = 0}^{\infty} \frac{1}{{(1 + 2 \frac{k}{n})}^{2}}

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

(2)

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

= \frac{1}{2} (- \frac{1}{x + 1}) ∣_{0}^{\infty}

= \frac{1}{2}

(1) is obtained by realizing that

n + 2 k

is negligible in comparison to

{(n + 2 k)}^{2}

as n approaches

\infty

(2) uses the well-known identity

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RizerMix

Expert2022-01-27Added 656 answers

Using integration by parts, we obtain

\int_{0}^{1} \frac{n x^{n - 1}}{1 + x} d x = \frac{x^{n}}{1 + x} |_{0}^{1} + \int_{0}^{1} \frac{x^{n}}{(1 + x)^{2}} d x = \frac{1}{2} + r_{n}

where clearly

0 < r_{n} \leq \int_{0}^{1} x^{n} d x = \frac{1}{n + 1} \to 0

, as

n \to \infty

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