Question

Harmonic Series Question

Answer 1

shoot

Answer 2

rewrite \[\sum_{n=1}^{\infty} 1/2(n+1) \] as \[c1\sum_{j=1}^{\infty} 1/J + c2\]

Answer 3

my teacher left a hint saying use \[\sum_{n=1}^{\infty} a_n +1 = \sum_{j=1}^{\infty} a_j +a_1\]

Answer 4

Not much of a hint, unless j starts at 2 for the right-hand side

Answer 5

out of my league for the moment, though i'll scribble the rest of the night.....

Answer 6

Thanks anyways

Answer 7

I assume this is what you have
\[\sum_{n=1}^{\infty}\frac{1}{2(n+1)}\]
and you want to re-write it?

Answer 8

yeah, thats what the teacher wants me to do. Rewrite it in the form of \[c_1 \sum_{j=1}^{\infty} 1/J + c_2\] . The second part of the question asks if it is convergent or divergent but I used the integral test to figure that out.

Answer 9

So the first series is 1/4, 1/6, 1/8, ..., yes?

Answer 10

that makes it a lot easier :)

Answer 11

yeah, you're right James.

Answer 12

One way to see how to do it is to write out a few terms:
\[= \frac{1}{2\cdot 2}+ \frac{1}{2\cdot 3}+ \frac{1}{2\cdot 4}+...\]

Answer 13

but starting at j=2 makes more sense

Answer 14

You can factor out the 1/2 from each term, right?

Answer 15

and what you have is almost the harmonic series, but it is missing the first term

Answer 16

yeah factoring out 1/2 would create a value for C_1 but I'm not sure what J is in the rewritten form, or C_2

Answer 17

In which case I'd just write it as

\[\frac{1}{2} \sum_{j=1}^{\infty} \frac{1}{j} - \frac{1}{2}\]

Answer 18

I ended up rewriting it as \[1/2 \sum_{n=1}^{\infty} 1/(n+1)\]

Answer 19

The converge or otherwise of that sum in j is a very important result.

Answer 20

I interpret that as
\[\sum_{j=1}^{\infty}\frac{1}{j}\]

Answer 21

It's the harmonic series. But it has an extra term, so subtract off 1/2
@JamesJ I am doing something wrong?

Answer 22

I'm wondering where the -1/2 came from

Answer 23

Yes, because there's no term 1/2 in the original series. That's why I subtracted it.

Answer 24

Compare the terms from the original series with my expression and you'll see they are identical

1/4, 1/6, 1/8, 1/10 , ....

Answer 25

As for the convergence, you have the harmonic series, which passes the integral test, but it diverges.

Answer 26

It fails the integral test ...

Answer 27

yeah, the series diverges