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Question Number 129869 by Bird last updated on 20/Jan/21

calculate lim_(n→+∞) Σ_(k=1) ^n (1/( (√((k+n)(k+1+n)))))

$${calculate}\:{lim}_{{n}\rightarrow+\infty} \sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{\left.\:\sqrt{\left({k}+{n}\right)\left({k}+\mathrm{1}+{n}\right.}\right)} \\ $$

Answered by mindispower last updated on 20/Jan/21

(1/((k+1+n)))≤(1/( (√(k+n)).(√(k+1+n))))≤(1/(k+n))...1 Σ(1/(k+n))=Σ_(k≤n) (1/n).(1/(1+(k/n)))=∫_0 ^1 (1/(1+x))=ln(2) Σ_(k≤n) ((1/(k+n))−(1/(k+1+n)))=Σ_(k≤n) (1/((k+n)(k+1+n)))=T T≤Σ_(k≤n) .(1/((1+n)(k+n)))≤Σ_(k≤n) (1/(n(n+1)))=(n/(n(n+1)))=(1/(n+1))→0 ⇒Σ_(k≤n) (1/((k+n)))−Σ_(k≤n) (1/((k+1+n)))→0 since Σ_(k≤n) (1/((k+n)))→ln(2)⇒Σ(1/(k+1+n))→ln(2) ⇒Σ(1/( (√(k+n.)).(√(k+1+n))))→ln(2) by using 1

$$ \\ $$$$\frac{\mathrm{1}}{\left({k}+\mathrm{1}+{n}\right)}\leqslant\frac{\mathrm{1}}{\:\sqrt{{k}+{n}}.\sqrt{{k}+\mathrm{1}+{n}}}\leqslant\frac{\mathrm{1}}{{k}+{n}}...\mathrm{1} \\ $$$$\Sigma\frac{\mathrm{1}}{{k}+{n}}=\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{{n}}.\frac{\mathrm{1}}{\mathrm{1}+\frac{{k}}{{n}}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}}={ln}\left(\mathrm{2}\right) \\ $$$$\underset{{k}\leqslant{n}} {\sum}\left(\frac{\mathrm{1}}{{k}+{n}}−\frac{\mathrm{1}}{{k}+\mathrm{1}+{n}}\right)=\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{\left({k}+{n}\right)\left({k}+\mathrm{1}+{n}\right)}={T} \\ $$$${T}\leqslant\underset{{k}\leqslant{n}} {\sum}.\frac{\mathrm{1}}{\left(\mathrm{1}+{n}\right)\left({k}+{n}\right)}\leqslant\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}=\frac{{n}}{{n}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{n}+\mathrm{1}}\rightarrow\mathrm{0} \\ $$$$\Rightarrow\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{\left({k}+{n}\right)}−\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{\left({k}+\mathrm{1}+{n}\right)}\rightarrow\mathrm{0} \\ $$$${since}\:\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{\left({k}+{n}\right)}\rightarrow{ln}\left(\mathrm{2}\right)\Rightarrow\Sigma\frac{\mathrm{1}}{{k}+\mathrm{1}+{n}}\rightarrow{ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\Sigma\frac{\mathrm{1}}{\:\sqrt{{k}+{n}.}.\sqrt{{k}+\mathrm{1}+{n}}}\rightarrow{ln}\left(\mathrm{2}\right)\:{by}\:{using}\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

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