Question-67996

Question Number 67996 by TawaTawa last updated on 03/Sep/19

Commented by mathmax by abdo last updated on 03/Sep/19

w e h a v e 1 ⩽ k ⩽ n \Rightarrow n^{2} + 1 ⩽ n^{2} + k ⩽ n^{2} + n \Rightarrow

\sqrt{n^{2} + 1} ⩽ \sqrt{n^{2} + k} ⩽ \sqrt{n^{2} + n} \Rightarrow \frac{1}{\sqrt{n^{2} + n}} ⩽ \frac{1}{\sqrt{n^{2} + k}} ⩽ \frac{1}{\sqrt{n^{2} + 1}} \Rightarrow

\frac{n}{\sqrt{n^{2} + n}} ⩽ \sum_{k = 1}^{n} \frac{1}{\sqrt{n^{2} + k}} ⩽ \frac{n}{\sqrt{n^{2} + 1}} w e h a v e

{l i m}_{n \to + \infty} \frac{n}{\sqrt{n^{2} + n}} = {l i m}_{n \to + \infty} \frac{n}{n \sqrt{1 + \frac{1}{n}}} = {l i m}_{n \to + \infty} \frac{1}{\sqrt{1 + \frac{1}{n}}} = 1

{l i m}_{n \to + \infty} \frac{n}{\sqrt{n^{2} + 1}} = {l i m}_{n \to + \infty} \frac{n}{n \sqrt{1 + \frac{1}{n^{2}}}} = {l i m}_{n \to + \infty} \frac{1}{\sqrt{1 + \frac{1}{n^{2}}}} = 1

\Rightarrow {l i m}_{n \to + \infty} a_{n} = 1 .

Commented by mathmax by abdo last updated on 03/Sep/19

b_{n} = \sum_{k = 1}^{n} \frac{1}{\sqrt{n + k}} w e h a v e 1 ⩽ k ⩽ n \Rightarrow n + 1 ⩽ n + k ⩽ 2 n \Rightarrow

\sqrt{n + 1} ⩽ \sqrt{n + k} ⩽ \sqrt{2 n} \Rightarrow \frac{1}{\sqrt{2 n}} ⩽ \frac{1}{\sqrt{n + k}} ⩽ \frac{1}{\sqrt{n + 1}} \Rightarrow

\frac{1}{\sqrt{n + k}} ⩾ \frac{1}{\sqrt{2 n}} b u t {l i m}_{n \to + \infty} \frac{1}{\sqrt{2 n}} = + \infty \Rightarrow {l i m}_{n \to + \infty} b_{n} = + \infty

c_{n} = \frac{1}{n} \sum_{k = n}^{2 n} \frac{1}{k} c h a n g e m e n t o f i n d i c e k - n = i g i v e

c_{n} = \frac{1}{n} \sum_{i = 0}^{n} \frac{1}{n + i} = \frac{1}{n} {\frac{1}{n} \sum_{i = 0}^{n} \frac{1}{1 + \frac{i}{n}}} w e h a v e

{l i m}_{n \to + \infty} \frac{1}{n} \sum_{i = 0}^{n} \frac{1}{1 + \frac{i}{n}} = \int_{0}^{1} \frac{d x}{1 + x} = {[l n ∣ 1 + x ∣]}_{0}^{1} = l n (2) \Rightarrow

{l i m}_{n \to + \infty} c_{n} = 0

Commented by TawaTawa last updated on 03/Sep/19

God bless you sir

Commented by Abdo msup. last updated on 04/Sep/19

y o u a r e w e l c o m e .

Answered by mind is power last updated on 03/Sep/19

a_{n} = \sum_{k = 1}^{n} \frac{1}{\sqrt{n^{2} + k}}

\forall k \in [1, n]

\frac{1}{\sqrt{n^{2} + n}} ⩽ \frac{1}{\sqrt{n^{2} + k}} ⩽ \frac{1}{\sqrt{n^{2} + 1}}

\Rightarrow \sum_{k = 1}^{n} \frac{1}{\sqrt{n^{2} + n}} ⩽ a_{n} ⩽ \sum_{k = 1}^{n} \frac{1}{\sqrt{n^{2} + 1}}

\Rightarrow \frac{n}{\sqrt{n^{2} + n}} ⩽ a_{n} ⩽ \frac{n}{\sqrt{n^{2} + 1}}

\Rightarrow {l i m a}_{n} = 1

\sum_{k = 1}^{n} \frac{1}{\sqrt{n + k}} = b_{n}

\forall k \frac{1}{\sqrt{n + k}} ⩾ \frac{1}{\sqrt{n + n}} = \frac{1}{\sqrt{2 n}}

\Rightarrow \sum_{k = 1}^{n} \frac{1}{\sqrt{n + k}} ⩾ Σ \frac{1}{\sqrt{2 n}} = \frac{n}{\sqrt{2 n}} = \sqrt{\frac{n}{2}}

\Rightarrow b_{n} \to + \infty

c_{n} = \frac{1}{n} \sum_{n}^{2 n} \frac{1}{k} = \frac{1}{n_{}} \sum_{k = n}^{k = 2 n} \frac{1}{k}

w e s h o w \sum_{k = n}^{k = 2 n} \frac{1}{k_{}} c v \Rightarrow c_{n} \to 0

\sum_{n}^{2 n} \frac{1}{k} = \sum_{k = 0}^{n} \frac{1}{2 n - k} = \frac{1}{n} \sum_{k = 0}^{n} \frac{1}{2 - \frac{k}{n}}

l e t f (x) = \frac{1}{2 - x}

\Rightarrow \frac{1}{n} \sum_{k = 0}^{n} \frac{1}{2 - \frac{k}{n}} = \frac{1}{n} \sum_{k = 0}^{n} f (\frac{k}{n}) = \int_{0}^{1} f (x) d x r e i m a n

\int_{0}^{1} f (x) d x = \int_{0}^{1} \frac{1}{2 - x} d x = {[- l n (2 - x)]}_{0}^{1} = l n (2)

\Rightarrow \sum_{k = n}^{2 n} \frac{1}{k} \to l n (2)

\Rightarrow \frac{1}{n} \sum_{k = n}^{k = 2 n} \to 0

Commented by TawaTawa last updated on 03/Sep/19

God bless you sir