How do I prove that (vn), defined byvn=1n2+1+1n2+2+⋯+1n2+n,is convergent?

Find out the answer to "How do I prove that (vn), defined byvn=1n2+1+1n2+2+⋯+1n2+n,is..."

Beckham Short

2022-04-13

How do I prove that

(v_{n})

, defined by

v_{n} = \frac{1}{\sqrt{n^{2} + 1}} + \frac{1}{\sqrt{n^{2} + 2}} + \dots + \frac{1}{\sqrt{n^{2} + n}},

is convergent?

entreblogsmc2j

Beginner2022-04-14Added 10 answers

\frac{n}{\sqrt{n^{2} + n}} = \frac{1}{\sqrt{n^{2} + n}} + \frac{1}{\sqrt{n^{2} + n}} + \dots + \frac{1}{\sqrt{n^{2} + n}} \leq

\frac{1}{\sqrt{n^{2} + 1}} + \frac{1}{\sqrt{n^{2} + 2}} + \dots + \frac{1}{\sqrt{n^{2} + n}}

\leq \frac{1}{\sqrt{n^{2}}} + \frac{1}{\sqrt{n^{2}}} + \dots + \frac{1}{\sqrt{n^{2}}} = \frac{n}{\sqrt{n^{2}}} = 1

The inequalities hold since bigger denominators make smaller fractions and obviously:

n^{2} + n \geq n^{2} + i \geq n^{2}

for

i = 1, \dots, n

Next,

lim_{n \to \infty} \frac{n}{\sqrt{n^{2} + n}} = 1

. So your summation is squeezed between 1 and 1. Thus its limit is 1.

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