Question

Show that
1/(r+1)(r+2)-1/r(r+1)=-2/r(r+1)(r+2)

Answer 1

\[\large \left(1/2 \sum \limits_{r=1}^n \dfrac{1}{r} - \dfrac{1}{r+1}\right) -\left(1 /2 \sum \limits_{r=1}^n \ \dfrac{1}{r+1} -\dfrac{1}{r+2} \right)\]

everything clear till this step ? :)

Answer 2

ys..

Answer 3

good, remembe the telescoping formula which we have derived in earlier post ?

Answer 4

Worth Repeating...

Warning!!! No, you CANNOT find the sum of the Harmonic Series. It is Divergent.

Divergence does NOT mean we are just not clever enough to find the sum, it means there ISN'T one.

Up above, where ganshie8 split the sum into two, separate sums -- that is NOT okay, unless both the new sums are convergent.

If you try this with the Harmonic Series, you will split your divergent sum into two other divergent sums. This is not helpful.

Answer 5

I was wondering... can we do the following?

\(\large\frac{1}{(r+1)(r+2)}-\frac{1}{r(r+1)}\)

=\(\large\frac{r}{r(r+1)(r+2)}-\frac{r+2}{r(r+1)(r+2)}\)

=\(\large \frac{r-(r+2)}{r(r+1)(r+2)}\)

=\(\large -\frac{2}{r(r+1)(r+2)}\)

Or did the question ask for a series solution?

Answer 6

I'm nt sure
The given answer is 1/2[1/2- 1/(n+1)(n+2)]
@mathmate

Answer 7

convergence/divergence should not matter when working partial sum, right ? @tkhunny

Answer 8

yes @mathmate after proving that, the question asks to find the partial sum

Answer 9

@ganeshie8
Oh... thank you.
Also, agree that divergence does not matter with partial sums.

Answer 10

This's my frien's working
Anyway, let's continue @ganeshie8