Question

Prove that this sequence converges:

s=1+1/2+1/3+...+1/n-ln(n)

Answer 1

\[S=1+(1/2)+(1/3)+...+(1/n)-\ln(n)\]

Answer 2

isnt that a series tho?

Answer 3

It is :D
But the series can't converge if the sequence doesn't converge, right? :D

Answer 4

those are state secrets that must never be revealed .... or at least they are details that I never remember :)

Answer 5

i know 1/n is divergent on it own; but that doesnt help

Answer 6

@amistre64
that, and I don't know what "ln" is doing in this series, it seems out of place...
is it though?

Answer 7

it does seem rather "odd" lol

Answer 8

Well, the serie should be \[(\sum_{k=1}^{n} 1/k) - \ln(n)\] doesn't?

Answer 9

It will diverge...
It's what you call the harmonic series, unless I'm missing something here
What's n supposed to be? :)

Answer 10

Wait a sec, this isn't an infinite series at all!
Then doesn't that mean that for sure, the series will converge? :)

Answer 11

That's the problem, it must converge, cause the limit of this sequence is Euler constant

Answer 12

the limit as n to inf i believe

Answer 13

can you define a rule for the partial sums? and then work that

Answer 14

What is n, though?
Is it a random number?
Or perhaps, you let n approach infinity?

Answer 15

s1 = 1
s2 = 3/2 - ln(2)
s3 = 3/2 - ln(2) +1/3-ln(3)
= 11/6 -ln(2/3)

hmmm

i think they forgot the limit on the front

Answer 16

n takes the last value for n in the serie, as S=1+1/2+1/3-ln(3) when k goes from 1 to 3

Answer 17

and -ln(n) = ln(1/n)

Answer 18

Are you tasked to prove that it converges, or
to prove that it converges to e?

Answer 19

Prove that the sequence converges

Answer 20

good luck, work beckons ....

Answer 21

seq to me suggests that the seq of partial sums as n to inf

Answer 22

Show that it's a cauchy sequence:
\[S_{n-1}-S_n=ln(n)-ln(n-1)-\frac{1}{n}\rightarrow0\]

Answer 23

Every real number cauchy sequence converges.

Answer 24

If I remember right, Sn is decreasing (or increasing?) and bounded, therefore, by Monotone Convergence Theorem, Sn converges.

Answer 25

Yes, I was working on that, I prove it's increasing but what about the bounds?

Answer 26

Doesn't the monotone convergence theorem only apply to the convergence of the sequence?

Answer 27

And the sequence in question is \(S_n=H_n-\ln(n)\), where Hn is the nth harmonic number.
@knock I think a geometric point of view may help. Consider the function \(f(x)=\frac{1}{x}\).

Answer 28

Ok

Answer 29

Why am I being ignored :p, I already proved it.

Answer 30

@blockcolder I think only the (1/k) is in the sigma;
ln (n) is subtracted from the sum, right?

Answer 31

Yeah, and...?

Answer 32

and nothing
I was just clarifying that ln(n) is not part of the sequence, only that it's subtracted from the series.

Answer 33

The series does converge, for sure (since it's finite)
When you say "its limit" what do you mean by it?
Do you mean as n goes to infinity?

Answer 34

The limit of this sequence as n goes to infinity is a number, called Euler constant and represented by leter gamma

Answer 35

But I just want to prove that the sequence converges

Answer 36

Ok here's a solution I found http://www.proofwiki.org/wiki/Euler-Maclaurin_Summation_Formula

Answer 37

I think if all you want to do is prove convergence, then, the integral test should suffice?

Answer 38

Yes but, as ln(n) it's not in the serie, can we use integral test?