Question

How do you prove that two series are identical?

Answer 1

@CGGURUMANJUNATH

Answer 2

@phi

Answer 3

?

Answer 4

I am just not sure what qualifies as sufficient proof to show that two series are the same

Answer 5

if each term in the series corresponds to the other series they are identical

or are you asking, how do you show their limits are the same ?

Answer 6

Well my book gave me a series, and then it said "tell whether the series is the same as" and then it gave me another series, and I am not sure how to prove that two series will be identical

Answer 7

can you post the question ?

Answer 8

\[\sum_{n=1}^{\infty}(-\frac{ 1 }{ 2 })^{n-1}\]

Answer 9

is that the same as \[\sum_{n=0}^{\infty}(-1)^{n}(\frac{ 1 }{ 2 })^{n}\]

Answer 10

And I know that they are the same, but I'm not sure how to PROVE it

Answer 11

we try to write them so we get the same formula for both.
for example, the first one can be rewritten to start at n=0
\[ \sum_{n=1}^{\infty}(-\frac{ 1 }{ 2 })^{n-1} = \sum_{m=0}^{\infty}(-1\cdot \frac{ 1 }{ 2 })^{m} \]
and then
\[\sum_{m=0}^{\infty}(-1)^m\cdot \left(\frac{ 1 }{ 2 }\right)^{m} \]

Answer 12

So you think that I can just say \[\sum_{n=1}^{\infty}(-\frac{ 1 }{ 2 })^{n-1}=\sum_{n=0}^{\infty}(-\frac{ 1 }{ 2 })^{n}\] ?

Answer 13

notice I let:
m= n-1
(so it follows that m+1)
I replace n-1 with m in the body
in the summation, I replace n with m+1 to get
m+1= 1
and simplified to get m=0 as the lower limit

Answer 14

it would be better to then factor out (-1)^n so that you get an exact match to the second summation

Answer 15

Aha. I see. So you replaced the m with m+1 and then you can simplify to show that the series are identical. Ok

Answer 16

Thank you!!!