El_Arrow
  • El_Arrow
use comparison or limit comparison test
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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El_Arrow
  • El_Arrow
\[\sum_{k=1}^{\infty} \frac{ 3\sqrt{k} +2 }{ \sqrt{k ^{3}+3k ^{2}+1} }\]
El_Arrow
  • El_Arrow
help me please i dont know what to do
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 3\sqrt{k} }{\sqrt{k^3+3k^2+1}}>\sum_{ n=1 }^{ \infty }\frac{ 3\sqrt{k} }{\sqrt{k^3+3k^3+k^3}}}\)

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SolomonZelman
  • SolomonZelman
simplify the right side and if the right side diverges then wouldn't the left side diverge as well?
El_Arrow
  • El_Arrow
yeah
SolomonZelman
  • SolomonZelman
ok, what do you get for the right side?
El_Arrow
  • El_Arrow
\[\frac{ 3\sqrt{k} }{ \sqrt{5k ^{3}} }\]
SolomonZelman
  • SolomonZelman
and that simplifies to what ?
El_Arrow
  • El_Arrow
i think it simplifies to 1/5k^(1/6)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{ 3\sqrt{k} }{\sqrt{k^3+3k^3+k^3}}=}\) \(\large\color{black}{ \displaystyle \frac{ 3\sqrt{k} }{\sqrt{5k^3}}=}\) \(\large\color{black}{ \displaystyle \frac{ 3\sqrt{k} }{\sqrt{5}\sqrt{k^3}}=}\) \(\large\color{black}{ \displaystyle \frac{3}{\sqrt{5}} \times \frac{ k^{1/2} }{k^{3/2}}=}\) \(\large\color{black}{ \displaystyle \frac{3}{\sqrt{5}} \times \frac{1 }{k^1}.}\)
SolomonZelman
  • SolomonZelman
Therefore: \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{ 3\sqrt{k} }{\sqrt{k^3+3k^3+k^3}}=\frac{3}{\sqrt{5}}\sum_{ n=1 }^{ \infty }\frac{ 1 }{k}}\)
SolomonZelman
  • SolomonZelman
What do we know abotu harmonic series ?
El_Arrow
  • El_Arrow
it diverges
SolomonZelman
  • SolomonZelman
And and if \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 3\sqrt{k} }{\sqrt{k^3+3k^2+1}}>\sum_{ n=1 }^{ \infty }\frac{ 3\sqrt{k} }{\sqrt{k^3+3k^3+k^3}}}\) then it is same as: \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 3\sqrt{k} }{\sqrt{k^3+3k^2+1}}>\frac{3}{\sqrt{5}}\sum_{ n=1 }^{ \infty }\frac{ 1 }{k}}\)
El_Arrow
  • El_Arrow
so the whole series diverges
SolomonZelman
  • SolomonZelman
I am using n as index, my error, but you get what I am saying
SolomonZelman
  • SolomonZelman
Yes, so your series diverges
El_Arrow
  • El_Arrow
could you help me with one more its \[\frac{ 1 }{ 1+e ^{^{k}} }\]
El_Arrow
  • El_Arrow
would you use the harmonic series on this one too?
SolomonZelman
  • SolomonZelman
So, \(\large\color{black}{ \displaystyle \sum_{ k=1 }^{ \infty } ~\frac{1}{1+e^k}}\) ?
El_Arrow
  • El_Arrow
yes
SolomonZelman
  • SolomonZelman
Ok, tell me if the following coverges or not: \(\large\color{black}{ \displaystyle \sum_{ k=1 }^{ \infty } ~\frac{1}{e^k}}\)
SolomonZelman
  • SolomonZelman
converges*
El_Arrow
  • El_Arrow
why?
SolomonZelman
  • SolomonZelman
Because this series is greater than YOUR series (since you are dividing by a smaller value, without the +1), and if this series converges (which you should know that it does). THEN your series which is smaller (because you are dividing by a bigger value) would also do what?
El_Arrow
  • El_Arrow
it would also converge
SolomonZelman
  • SolomonZelman
yes
El_Arrow
  • El_Arrow
thank you sir
SolomonZelman
  • SolomonZelman
Anytime!

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