Question

Why is this statement true?
\[\sum_{n=0}^{\infty} nx^{n-1} = \sum_{n=0}^{\infty} (n+1)x^n\]

Answer 1

did you forget to change the limits?
do you mean why is this statement true:
\[\sum_{n=1}^{\infty} nx^{n-1}=\sum_{n=0}^{\infty} (n+1) x^{n}\]

Answer 2

this is because
n=1 means n-1=0
say we let k=n-1
then we have k=0 since n-1=0

and if n=infty then k=n-1 is also infinity

so
\[\sum_{n=1}^{n=\infty} n x^{n-1} \\ \sum_{n-1=1-1}^{n-1=\infty-1} n x^{n-1} \\ \text{ noticed I subtracted one on both sides of the upper and lower limit of the sum } \\ n=(n-1)+1 \\ \sum_{n-1=0}^{n-1=\infty } ((n-1)+1)x^{n-1} \\ \text{ now \let } n-1=k \\ \text{ then you have } \\ \sum_{k=0}^{k=\infty}(k+1)x^{k} \\ \sum\limits_{k=0}^{\infty} (k+1)x^{k}\]
can you change k to any letter you want
maybe back to n if you prefer
\[\sum_{n=0}^{\infty} (k+1)x^{k}\]

also you shouldn't really use n=infty since infty isn't really a number
I was just doing it because I thought the explanation would come easier
It was limit... if n goes to infinity then so does n-1

Answer 3

i'd understand it if it were just a matter of reducing the index.. however this is the problem im working on and on the 2nd line it does that and im not sure why/how
http://math.stackexchange.com/questions/345538/power-series-representation-of-frac1x1-x2

Answer 4

Oh yes I overlooked that earlier. Those are also equal. You can show this by writing the first term of the series plus rest of series and indexing that rest part. What I mean is \[\sum_{n=0}^{\infty} n x^{n-1}=0x^{0-1}+\sum_{n=1}^{\infty} n x^{n-1}\] now do that indexing thing above I showed