Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 3564 by Yozzii last updated on 15/Dec/15

Test for convergence:  (1) Σ_(n=10) ^∞ (2^(ln(lnn)) /(nlnn))  (2)Σ_(n=2) ^∞ (1/(n(lnn)^p )) (two cases of p to look at)  (3)Σ_(n=2) ^∞ (((−1)^n (√n))/(lnn))  (4)Σ_(n=1) ^∞ ((10^n n)/((2n+1)!))  (5)Σ_(n=1) ^∞ ((n!)/n^n )  This post is my attempt to adhere to  the common agreement on infinite  series studying. Have fun!

$${Test}\:{for}\:{convergence}: \\ $$$$\left(\mathrm{1}\right)\:\underset{{n}=\mathrm{10}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{\mathrm{ln}\left(\mathrm{ln}{n}\right)} }{{n}\mathrm{ln}{n}} \\ $$$$\left(\mathrm{2}\right)\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{ln}{n}\right)^{\mathrm{p}} }\:\left(\mathrm{two}\:\mathrm{cases}\:\mathrm{of}\:\mathrm{p}\:\mathrm{to}\:\mathrm{look}\:\mathrm{at}\right) \\ $$$$\left(\mathrm{3}\right)\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \sqrt{{n}}}{\mathrm{ln}{n}} \\ $$$$\left(\mathrm{4}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{10}^{{n}} {n}}{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$\left(\mathrm{5}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!}{{n}^{{n}} } \\ $$$${This}\:{post}\:{is}\:{my}\:{attempt}\:{to}\:{adhere}\:{to} \\ $$$${the}\:{common}\:{agreement}\:{on}\:{infinite} \\ $$$${series}\:{studying}.\:{Have}\:{fun}! \\ $$

Commented by prakash jain last updated on 15/Dec/15

Find sum or only check for convergence?

$$\mathrm{Find}\:\mathrm{sum}\:\mathrm{or}\:\mathrm{only}\:\mathrm{check}\:\mathrm{for}\:\mathrm{convergence}? \\ $$

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!

$${Do}\:{as}\:{you}\:{please},\:{but}\:{I}\:{only}\:{asked}\: \\ $$$${about}\:{convergence}.\:{If}\:{a}\:{question} \\ $$$${sparks}\:{further}\:{questions}\:{then}\:{fantastic}! \\ $$

Answered by prakash jain last updated on 15/Dec/15

5. a_n =((n!)/n^n ), a_(n+1) =(((n+1)!)/((n+1)^(n+1) ))  (a_(n+1) /a_n )=(((n+1)!n^n )/(n!(n+1)^(n+1) ))=(((n+1)n^n )/((n+1)(n+1)^n ))=((n/(1+n)))^n   lim_(n→∞) ((n/(1+n)))^n =lim_(x→0) (((1/x)/(1+1/x)))^(1/x) =lim_(x→0) ((1/(1+x)))^(1/x)   y=((1/(1+x)))^(1/x) ⇒ln y=−((ln (1+x))/x)  lim_(x→0)  ln y=lim_(x→0) −((1/(1+x))/1)=−1  lim_(x→0)  ((1/(1+x)))^(1/x) =(1/e)  (1/e)<1, hence by ratio test series converges.

$$\mathrm{5}.\:{a}_{{n}} =\frac{{n}!}{{n}^{{n}} },\:{a}_{{n}+\mathrm{1}} =\frac{\left({n}+\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} } \\ $$$$\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\frac{\left({n}+\mathrm{1}\right)!{n}^{{n}} }{{n}!\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }=\frac{\left({n}+\mathrm{1}\right){n}^{{n}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)^{{n}} }=\left(\frac{{n}}{\mathrm{1}+{n}}\right)^{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}}{\mathrm{1}+{n}}\right)^{{n}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}/{x}}{\mathrm{1}+\mathrm{1}/{x}}\right)^{\mathrm{1}/{x}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\mathrm{1}/{x}} \\ $$$${y}=\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\mathrm{1}/{x}} \Rightarrow\mathrm{ln}\:{y}=−\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{ln}\:{y}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\frac{\mathrm{1}}{\mathrm{1}+{x}}}{\mathrm{1}}=−\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\mathrm{1}/{x}} =\frac{\mathrm{1}}{{e}} \\ $$$$\frac{\mathrm{1}}{{e}}<\mathrm{1},\:\mathrm{hence}\:\mathrm{by}\:\mathrm{ratio}\:\mathrm{test}\:\mathrm{series}\:\mathrm{converges}. \\ $$

Answered by prakash jain last updated on 15/Dec/15

4. a_n =((10^n n)/((2n+1)!)) ⇒  (a_(n+1) /a_n )=((10(n+1))/(n(2n+2)(2n+3)))  lim_(n→∞)  ((10(n+1))/(n(2n+2)(2n+3)))=0  series converges

$$\mathrm{4}.\:{a}_{{n}} =\frac{\mathrm{10}^{{n}} {n}}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\Rightarrow\:\:\frac{{a}_{{n}+\mathrm{1}} }{{a}_{{n}} }=\frac{\mathrm{10}\left({n}+\mathrm{1}\right)}{{n}\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{10}\left({n}+\mathrm{1}\right)}{{n}\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}=\mathrm{0} \\ $$$$\mathrm{series}\:\mathrm{converges} \\ $$

Answered by prakash jain last updated on 15/Dec/15

3. a_n =(((−1)^n (√n))/(ln n))  Ratio test, r_n = ((√(n+1))/(ln (n+1)))×((ln n)/(√n))  =(√(1+(1/n)))×((ln n)/(ln n+1))  lim_(n→0) r_n =1 (inconclusive)  lim_(n→∞) ∣a_n ∣=lim_(n→∞)  ((√n)/(ln n))=(∞/∞)  lim_(n→∞) ∣a_n ∣=lim_(n→∞)  ((√n)/(ln n))=lim_(n→∞) (n/(2(√n)))=∞  series diverges.

$$\mathrm{3}.\:{a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} \sqrt{{n}}}{\mathrm{ln}\:{n}} \\ $$$$\mathrm{Ratio}\:\mathrm{test},\:{r}_{{n}} =\:\frac{\sqrt{{n}+\mathrm{1}}}{\mathrm{ln}\:\left({n}+\mathrm{1}\right)}×\frac{\mathrm{ln}\:{n}}{\sqrt{{n}}} \\ $$$$=\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}}×\frac{\mathrm{ln}\:{n}}{\mathrm{ln}\:{n}+\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}{r}_{{n}} =\mathrm{1}\:\left(\mathrm{inconclusive}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\mid{a}_{{n}} \mid=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{n}}}{\mathrm{ln}\:{n}}=\frac{\infty}{\infty} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\mid{a}_{{n}} \mid=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{n}}}{\mathrm{ln}\:{n}}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{\mathrm{2}\sqrt{{n}}}=\infty \\ $$$$\mathrm{series}\:\mathrm{diverges}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com