Question

For what values of x does the series 1+2^x+3^x+4^x+...+n^x+... converge? (Answer: x<-1)

Answer 1

my guess is \(x<1\)

Answer 2

But the answer says x<-1, are you sure your answer is right? Can you show the work?

Answer 3

oh my guess was wrong, sorry ignore that one

Answer 4

\[\large\begin{array}{rcl}
\left|\lim_{n\rightarrow\infty}\frac{(n+1)^x}{n^x}\right|&<&1\\
\left(\lim_{n\rightarrow\infty}\frac{(n+1)^x}{n^x}\right)^2&<&1\\
\lim_{n\rightarrow\infty}\frac{(n+1)^{2x}}{n^{2x}}&<&1\\
\lim_{n\rightarrow\infty}[\ln(n+1)^{2x}-\ln n^{2x}]&<&0\\
\lim_{n\rightarrow\infty}[2x\ln(n+1)-2x\ln n]&<&0\\
\lim_{n\rightarrow\infty}2x[\ln(n+1)-\ln n]&<&0\\
\end{array}\]Wait this seems be not the approach correct...

Answer 5

Forgive me English of mine

Answer 6

@satellite73 you think something?

Answer 7

@Loser66 and about you?

Answer 8

i can't seem to figure out how to get the limit less than 1

Answer 9

Maybe the test wrong I using?

Answer 10

i think the log is the way to go

Answer 11

@Idealist What think-you?

Answer 12

\[(\frac{n+1}{n})^x<1\]
\[x\ln(n+1)-\ln(x)<0\] but i keep going round in circles

Answer 13

But same is that of mine?

Answer 14

\[x\ln(\frac{n+1}{n})<0\] yes same

Answer 15

And limit approach zero?

Answer 16

Gives you 0x<0?

Answer 17

That not makes sense...

Answer 18

having a brain fart
i think the key is solving
\[x\ln(\frac{n+1}{n})<0\] for \(x\)

Answer 19

But that gives x*0<0 which not makes sense...

Answer 20

which wolfram tells me the solution is \(x<-1\)

Answer 21

interesting

Answer 22

But what did wolfram do?

Answer 23

dunno i must be screwing up somewhere

Answer 24

What do you mean by false?

Answer 25

if \(x=-1\) you get \(\sum\frac{1}{n}\) which is well known to diverge

Answer 26

google "harmonic series" for more info :)

Answer 27

if \(x<-1\) say \(x=-(1+\epsilon)\) you get
\[\sum\frac{1}{n^{1+\epsilon}}\] where \(\epsilon>1\) and this converges, probably easiest to use the integral test

Answer 28

i meant where \(\epsilon>0\)

Answer 29

But how do we know that it converges if x>-1?

Answer 30

diverges*

Answer 31

if it diverges at \(x=-1\) it certainly diverges for \(x>-1\)

Answer 32

in any case if \(x=-1+\epsilon\) where \(\epsilon>0\) then you get
\[\sum\frac{1}{n^{1-\epsilon}}\] and that diverges again by the integral test

Answer 33

got off on the wrong foot by thinking of ratio test, rather than harmonic series etc

Answer 34

like i said, brain fart

Answer 35

HAHAHA!