Question

use comparison or limit comparison test

Answer 1

\[\sum_{k=1}^{\infty} \frac{ 3\sqrt{k} +2 }{ \sqrt{k ^{3}+3k ^{2}+1} }\]

Answer 2

help me please i dont know what to do

Answer 3

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 3\sqrt{k} }{\sqrt{k^3+3k^2+1}}>\sum_{ n=1 }^{ \infty }\frac{ 3\sqrt{k} }{\sqrt{k^3+3k^3+k^3}}}\)

Answer 4

simplify the right side and if the right side diverges then wouldn't the left side diverge as well?

Answer 5

yeah

Answer 6

ok, what do you get for the right side?

Answer 7

\[\frac{ 3\sqrt{k} }{ \sqrt{5k ^{3}} }\]

Answer 8

and that simplifies to what ?

Answer 9

i think it simplifies to 1/5k^(1/6)

Answer 10

\(\large\color{black}{ \displaystyle \frac{ 3\sqrt{k} }{\sqrt{k^3+3k^3+k^3}}=}\)

\(\large\color{black}{ \displaystyle \frac{ 3\sqrt{k} }{\sqrt{5k^3}}=}\)

\(\large\color{black}{ \displaystyle \frac{ 3\sqrt{k} }{\sqrt{5}\sqrt{k^3}}=}\)

\(\large\color{black}{ \displaystyle \frac{3}{\sqrt{5}} \times \frac{ k^{1/2} }{k^{3/2}}=}\)

\(\large\color{black}{ \displaystyle \frac{3}{\sqrt{5}} \times \frac{1 }{k^1}.}\)

Answer 11

Therefore:

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{ 3\sqrt{k} }{\sqrt{k^3+3k^3+k^3}}=\frac{3}{\sqrt{5}}\sum_{ n=1 }^{ \infty }\frac{ 1 }{k}}\)

Answer 12

What do we know abotu harmonic series ?

Answer 13

it diverges

Answer 14

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:

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 3\sqrt{k} }{\sqrt{k^3+3k^2+1}}>\frac{3}{\sqrt{5}}\sum_{ n=1 }^{ \infty }\frac{ 1 }{k}}\)

Answer 15

so the whole series diverges

Answer 16

I am using n as index, my error, but you get what I am saying

Answer 17

Yes, so your series diverges

Answer 18

could you help me with one more its \[\frac{ 1 }{ 1+e ^{^{k}} }\]

Answer 19

would you use the harmonic series on this one too?

Answer 20

So,

\(\large\color{black}{ \displaystyle \sum_{ k=1 }^{ \infty } ~\frac{1}{1+e^k}}\) ?

Answer 21

yes

Answer 22

Ok, tell me if the following coverges or not:

\(\large\color{black}{ \displaystyle \sum_{ k=1 }^{ \infty } ~\frac{1}{e^k}}\)

Answer 23

converges*

Answer 24

why?

Answer 25

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?

Answer 26

it would also converge

Answer 27

yes

Answer 28

thank you sir

Answer 29

Anytime!