Question Number 214485 by depressiveshrek last updated on 09/Dec/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}\centerdot\mathrm{7}}+…+\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{1}\right)}\right) \\ $$ Answered by mr W last updated on 10/Dec/24 $$\psi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=−\gamma+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}−\frac{\mathrm{1}}{\mathrm{3}}}\right) \\…
Question Number 214326 by mathlove last updated on 05/Dec/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}}]{{x}}−\sqrt[{\mathrm{6}}]{{x}}}{\:\sqrt[{\mathrm{4}}]{{x}}+\sqrt[{\mathrm{6}}]{{x}}}=? \\ $$ Answered by malwan last updated on 05/Dec/24 $${put}\:{y}={x}^{\mathrm{12}} \\ $$$$\underset{{y}\rightarrow\infty} {{lim}}\:\frac{{y}^{\mathrm{3}} −{y}^{\mathrm{2}}…
Question Number 214258 by efronzo1 last updated on 03/Dec/24 $$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} −\mathrm{x}^{\mathrm{5}} }−\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} +\mathrm{5x}^{\mathrm{5}} }\:=? \\ $$ Answered by issac last updated on 03/Dec/24 $$−\mathrm{1}…
Question Number 214216 by MathematicalUser2357 last updated on 01/Dec/24 $${a}_{{n}} ={a}_{{n}−\mathrm{1}} +\frac{{x}−{a}_{{n}−\mathrm{1}} }{{t}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} =? \\ $$ Answered by mr W last updated…
Question Number 214181 by golsendro last updated on 30/Nov/24 $$\:\:\:\:\underbrace{\:} \\ $$ Answered by efronzo1 last updated on 30/Nov/24 $$\:\:\:\cancel{\nprec} \\ $$ Terms of Service…
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Question Number 213991 by Spillover last updated on 23/Nov/24 Answered by MathematicalUser2357 last updated on 28/Nov/24 $$\mathrm{Then}\:\mathrm{K}\:\mathrm{is}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{r}} \centerdot\underset{{i}=\mathrm{1}} {\overset{{r}−\mathrm{1}} {\prod}}\left(\mathrm{1}−\mathrm{2}{i}\right)}{\mathrm{2}^{\mathrm{3}{r}+\mathrm{1}} \centerdot\left(\mathrm{2}{r}+\mathrm{1}\right)\centerdot{r}!} \\ $$…
Question Number 213744 by efronzo1 last updated on 15/Nov/24 Answered by golsendro last updated on 15/Nov/24 $$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{4x}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} +\left(\mathrm{1}−\mathrm{4x}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{1}+\mathrm{4x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{3}\left(\mathrm{1}−\mathrm{4x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}}…
Question Number 213642 by universe last updated on 12/Nov/24 Answered by Berbere last updated on 12/Nov/24 $${a}_{{n}} =\underset{{N}\rightarrow\infty} {\mathrm{lim}}\underset{{k}={n}} {\overset{{N}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} };\forall{k}\geqslant{n}>\mathrm{1}\:\:{k}\left({k}−\mathrm{1}\right)\:\leqslant{k}^{\mathrm{2}} \leqslant{k}\left({k}+\mathrm{1}\right)\Rightarrow \\ $$$$\underset{{N}\rightarrow\infty}…
Question Number 213548 by universe last updated on 08/Nov/24 $$\mathrm{0}<{c}<\mathrm{1}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{recursive}\:\mathrm{sequence} \\ $$$$\left\{{a}_{{n}} \right\}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{setting}\: \\ $$$$\:\mathrm{a}_{\mathrm{1}\:} =\:\frac{\mathrm{c}}{\mathrm{2}}\:\:,\:{a}_{\mathrm{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{c}+\mathrm{a}_{\mathrm{n}} ^{\mathrm{2}} \right)\:\:\mathrm{for}\:\mathrm{n}\in\:\mathbb{N} \\ $$$$\mathrm{monotonic}\:\mathrm{and}\:\mathrm{convergent} \\ $$ Answered by…