Test-for-convergence-1-n-10-2-ln-lnn-nlnn-2-n-2-1-n-lnn-p-two-cases-of-p-to-look-at-3-n-2-1-n-n-lnn-4-n-1-10-n-n-2n-1-5-n-1-

Question Number 3564 by Yozzii last updated on 15/Dec/15
$${Test}\:{for}\:{convergence}: \\$$$$\left(\mathrm{1}\right)\:\underset{{n}=\mathrm{10}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{\mathrm{ln}\left(\mathrm{ln}{n}\right)} }{{n}\mathrm{ln}{n}} \\$$$$\left(\mathrm{2}\right)\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{ln}{n}\right)^{\mathrm{p}} }\:\left(\mathrm{two}\:\mathrm{cases}\:\mathrm{of}\:\mathrm{p}\:\mathrm{to}\:\mathrm{look}\:\mathrm{at}\right) \\$$$$\left(\mathrm{3}\right)\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \sqrt{{n}}}{\mathrm{ln}{n}} \\$$$$\left(\mathrm{4}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{10}^{{n}} {n}}{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\$$$$\left(\mathrm{5}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!}{{n}^{{n}} } \\$$$${This}\:{post}\:{is}\:{my}\:{attempt}\:{to}\:{adhere}\:{to} \\$$$${the}\:{common}\:{agreement}\:{on}\:{infinite} \\$$$${series}\:{studying}.\:{Have}\:{fun}! \\$$
Commented by prakash jain last updated on 15/Dec/15
$$\mathrm{Find}\:\mathrm{sum}\:\mathrm{or}\:\mathrm{only}\:\mathrm{check}\:\mathrm{for}\:\mathrm{convergence}? \\$$
Commented by Yozzii last updated on 15/Dec/15
$${Do}\:{as}\:{you}\:{please},\:{but}\:{I}\:{only}\:{asked}\: \\$$$${about}\:{convergence}.\:{If}\:{a}\:{question} \\$$$${sparks}\:{further}\:{questions}\:{then}\:{fantastic}! \\$$
Answered by prakash jain last updated on 15/Dec/15
$$\mathrm{5}.\:{a}_{{n}} =\frac{{n}!}{{n}^{{n}} },\:{a}_{{n}+\mathrm{1}} =\frac{\left({n}+\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} } \\$$$$\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\frac{\left({n}+\mathrm{1}\right)!{n}^{{n}} }{{n}!\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }=\frac{\left({n}+\mathrm{1}\right){n}^{{n}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)^{{n}} }=\left(\frac{{n}}{\mathrm{1}+{n}}\right)^{{n}} \\$$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}}{\mathrm{1}+{n}}\right)^{{n}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}/{x}}{\mathrm{1}+\mathrm{1}/{x}}\right)^{\mathrm{1}/{x}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\mathrm{1}/{x}} \\$$$${y}=\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\mathrm{1}/{x}} \Rightarrow\mathrm{ln}\:{y}=−\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}} \\$$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{ln}\:{y}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\frac{\mathrm{1}}{\mathrm{1}+{x}}}{\mathrm{1}}=−\mathrm{1} \\$$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\mathrm{1}/{x}} =\frac{\mathrm{1}}{{e}} \\$$$$\frac{\mathrm{1}}{{e}}<\mathrm{1},\:\mathrm{hence}\:\mathrm{by}\:\mathrm{ratio}\:\mathrm{test}\:\mathrm{series}\:\mathrm{converges}. \\$$
Answered by prakash jain last updated on 15/Dec/15
$$\mathrm{4}.\:{a}_{{n}} =\frac{\mathrm{10}^{{n}} {n}}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\Rightarrow\:\:\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\frac{\mathrm{10}\left({n}+\mathrm{1}\right)}{{n}\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\$$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{10}\left({n}+\mathrm{1}\right)}{{n}\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}=\mathrm{0} \\$$$$\mathrm{series}\:\mathrm{converges} \\$$
Answered by prakash jain last updated on 15/Dec/15
$$\mathrm{3}.\:{a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} \sqrt{{n}}}{\mathrm{ln}\:{n}} \\$$$$\mathrm{Ratio}\:\mathrm{test},\:{r}_{{n}} =\:\frac{\sqrt{{n}+\mathrm{1}}}{\mathrm{ln}\:\left({n}+\mathrm{1}\right)}×\frac{\mathrm{ln}\:{n}}{\:\sqrt{{n}}} \\$$$$=\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}}×\frac{\mathrm{ln}\:{n}}{\mathrm{ln}\:{n}+\mathrm{1}} \\$$$$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}{r}_{{n}} =\mathrm{1}\:\left(\mathrm{inconclusive}\right) \\$$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\mid{a}_{{n}} \mid=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{n}}}{\mathrm{ln}\:{n}}=\frac{\infty}{\infty} \\$$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\mid{a}_{{n}} \mid=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{n}}}{\mathrm{ln}\:{n}}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{\mathrm{2}\sqrt{{n}}}=\infty \\$$$$\mathrm{series}\:\mathrm{diverges}. \\$$