Question

can someone show me how both series are diverges

Answer 1

\[\sum_{k=0}^{\infty} \frac{ \sqrt{2} }{ Ln^{10}k }\]

Answer 2

\[\sum_{k=1}^{\infty} k ^{1/k}\]

Answer 3

Did you copy this correctly $$\sum_{k=0}^{\infty} \frac{ \sqrt{2} }{ \ln^{10}(k) }$$I'm a little unsure about the exponent 10

Answer 4

$$\sum_{k=1}^{\infty} k ^{1/k}$$This series diverges. I just checked.

Answer 5

yea i just dont get why how to solve it

Answer 6

I'm confused about how this one looks. $$\sum_{k=0}^{\infty} \frac{ \sqrt{2} }{ Ln^{10}k }$$ can you draw it or take a screenshot.

Answer 7

sorry its \[\sqrt{k}\] at the top

Answer 8

Oh. Is there an Ln on the bottom?

Answer 9

yes

Answer 10

Ln^10 (k)

Answer 11

Is this it $$\sum_{k=0}^{\infty} \frac{ \sqrt{k} }{ \ln^{10}k }$$

Answer 12

Let's go over this one. This one is pretty straightforward:
$$ \sum_{k=1}^{\infty} k ^{1/k}$$
it diverges because of the divergence test.

Answer 13

Divergence Test for series:

If \( \lim_{k \to \infty} a_k \neq 0 \), then \( \sum_{k=0}^{\infty} a_k\) diverges.

Answer 14

The divergence test says, if the terms of the series don't go to zero there's no chance that the sum of the series can converge.

Answer 15

Clearly if you have a sequence with terms
1, 1, 1, 1, 1, ...
and then are asked to add these,
1 + 1 + 1 + 1 + ...
the sum diverges.

The terms of a series must approach zero if there is any chance for the sum of the terms to converge.
Of course that is not enough to get convergence. The terms could go to zero and yet the sum diverges.

Example : {1/n} starting from n =1

1, 1/2, 1/3, 1/4, 1/5, ...

the terms are going to zero but
1 + 1/2 + 1/3 + 1/4 + 1/5 + ... = \( \infty \)
$$ \sum_{n=1}^{\infty} \frac 1 n = \infty $$

But it could converge as well.

Example : {1/n^2} starting from n =1

1/1^2, 1/2^2, 1/3^2, 1/4^2, 1/5^2, ...
or
1/1 , 1/4, 1/9, 1/16, 1/25


the terms are going to zero and
1/1 + 1/4 + 1/9 + 1/16 + 1/25 + ... = \( \frac{\pi ^2}{6} \)

or to put it in symbols
$$ \sum_{n=1}^{\infty} \frac 1 {n^2} = \frac{\pi ^2}{6} $$

Answer 16

Is that clear?

Answer 17

thank you i get it now

Answer 18

\[n^{1/n}->1 \]

Answer 19

as n tends to infinity

Answer 20

So for your problem. $$\sum_{k=1}^{\infty} k ^{1/k}$$ all we have to show is that the sequence
\( \{ k^{1/k}\} : 1^{1/1} ,~ 2^{1/2}~, 3^{1/3} , ~ 4^{1/4}, ... \)

the terms do not approach zero.
i.e. we have to show the limit of \( \{ k^{1/k}\} \) as \(k \to \infty \) is non zero.

Answer 21

@jango_IN_DTOWN
thats correct. can you show that?

Answer 22

yeah

Answer 23

$$ \large
\lim_{k \rightarrow \infty} k^{1/k} =
\lim_{k \rightarrow \infty} e^{ \ln ( k^{1/k}) } =
\lim_{k \rightarrow \infty} e^{\frac{1}{k} \ln k} = e^{\lim_{k \rightarrow \infty} \frac{1}{k} \ln k}$$ Now use L'hopitals rule

Answer 24

here i used the fact that
$$ \large A = e^{ \ln A } $$

Answer 25

$$\Large
\lim_{k \rightarrow \infty} k^{1/k}
\\ \Large =\lim_{k \rightarrow \infty} e^{ \ln ( k^{1/k}) }
\\ \Large =
\lim_{k \rightarrow \infty} e^{\frac{1}{k} \ln k}
\\ \Large = e^{\lim_{k \rightarrow \infty} \frac{ \ln k } {k} }
\\ L.H. \Rightarrow \Large e^{\lim_{k \rightarrow \infty} \frac{ \frac 1 k } {1} }
\\ \Large = e^{\lim_{k \rightarrow \infty} \frac{ 1 } {k} }
\\ \Large = e^{ \frac{ 1 } { \infty } } \Large = e^{ 0 } = 1
$$

Answer 26

$$\large \sum_{k=0}^{\infty} \frac{ \sqrt{k} }{ \ln^{10}k }$$which i interpret to mean
$$ \large \sum_{k=0}^{\infty} \frac{ \sqrt{k} }{ \ln (k)^{10} }$$

Answer 27

@ jango_IN_DTOWN

in your proof why did you add an extra 1
$$ \large \sqrt{ \frac{2}{n-1} } < \epsilon \iff n > 1 + \frac{2}{\epsilon^2 } \color{red}{+ 1} $$ http://prntscr.com/cqwlv8

Answer 28

I suppose you can add 1 to get strict inequality.
it's a remarkable proof

$$\large n > 1 + \frac{2}{\epsilon^2 } \color{red}{+ 1}
\\ \, \\ \large \implies
\\ \, \\ \large \sqrt{ \frac{2}{n-1} }
= \sqrt{\frac{ 2 }{ 1 + \frac{2}{\epsilon^2 } \color{red}{+ 1} -1 } } <
\sqrt{\frac{ 2 }{ 1 + \frac{2}{\epsilon^2 } } } < \sqrt{\frac{ 2 }{ \frac{2}{\epsilon^2 } } } < \epsilon
$$

Answer 29

@jango_IN_DTOWN do you agree? or was there another reason

Answer 30

since n has to be an integer, you dont want fractional issues

Answer 31

oh... the term inside the [.] is not necessarily an integer, so i took the greatest integer function...now suppose we got n=5.2 then for n=5, the inequality will not be satisfied.
hence we need to add that 1. since if for n=5.2 the inequality is satisfied, it will be satisfied for n=6,7,.... but not 5

Answer 32

Nice.
I think i can make a proof along the same lines.

Let $$\large \rm h_n = ( n^{1/n} - 1 ) $$. We have
$$\rm \large n = ( 1 + h_n ) ^n
\\ \large = 1 + nh_n + n(n-1) \frac{(h_n)^2}{2!} ... + (h_n)^n > 1 + nh_n $$
From which it follows $$ \rm \large n > 1 + nh_n
\\ \large \implies h_n < \frac{n-1}{n}
\\ \large \implies h_n < 1 - \frac 1 n
$$ Then $$ \large \rm | h_n - 0 |= h_n < 1 - \frac 1 n < \epsilon~ \text{ if } n \ge N = \frac{1}{1- \epsilon } + 1 $$

Answer 33

right. I don't see any error, except the fact, the denominator is acceptable or not.

Answer 34

we can assume that epsilon is less than 1, to avoid division by zero

Answer 35

thats right. what about the other problem...@zaxoanl can you kindly provide a picture of the first problem?

Answer 36

I believe there's a problem with my proof. I tried to go back and plug it in but the inequalities don't work.
I can make a small adjustment

Answer 37

@zaxoanl Can you draw the other series using the draw tool?

Answer 38

I texted him to provide a picture.. For the second problem we got two solutions, first your method, then the method I provided. Now lets focus on the first problem.

Answer 39

@zaxoanl Does the first series really start at \(k=0\)?

Answer 40

I'll go ahead and assume the contrary, that the series begins for a more appropriate starting index:
\[\sum_{k=2}^\infty\frac{\sqrt k}{\ln^{10}k}=\sum_{k=2}^\infty\frac{\sqrt k}{(\ln k)^{10}}\]Consider the substitution \(n=\ln k\), so the series is equivalent to
\[\sum_{n\in L}\frac{e^{n/2}}{n^{10}}\]where \(L=\left\{\ln k~:~k\in\mathbb N\setminus\{1\}\right\}\). The limit test for divergence will show you that the series diverges, as
\[\begin{align*}
\frac{e^{n/2}}{n^{10}}&=\frac{1+\dfrac{n}{2^11!}+\dfrac{n^2}{2^22!}+\cdots+\dfrac{n^{10}}{2^{10}10!}+\dfrac{n^{11}}{2^{11}11!}+\cdots}{n^{10}}\\[1ex]
&=\frac{1}{n^{10}}+\frac{1}{2^11!n^9}+\frac{1}{2^22!n^8}+\cdots+\frac{1}{2^{10}10!}+\frac{n}{2^{11}11!}+\cdots\\[1ex]
&\to\infty
\end{align*}\]