# Difficulty to understand the value of a sum

• Jun 2nd 2011, 03:17 PM
Melsi
Difficulty to understand the value of a sum
I am trying to understant the formula below, it is part of bigger exercise but I am stack on this partiular one:

Sum from k=v+1 to n of (1/ (1/v+1)^(k-v-1)) = (1-(1/(v+1))^(n-v))/1-1/(v+1)
where v<n

This is in latex:
\sum_{k=v-1}^{n}\frac{1}{(v+1)^{k-v-1}}=\frac{1-(\frac{1}{v+1})^{n-v}}{1-\frac{1}{v+1}}

Sorry for latex, I am trying about 2 hours now.. The code produced from the buttons below doesn't work, the code in the latex forum doesn't work either, so I removed the math tags. Image upload does not work too, sorry!
• Jun 2nd 2011, 03:37 PM
Plato
Quote:

Originally Posted by Melsi
I am trying to understant the formula below, it is part of bigger exercise but I am stack on this partiular one:

Sum from k=v+1 to n of (1/ (1/v+1)^(k-v-1)) = (1-(1/(v+1))^(n-v))/1-1/(v+1)
where v<n

This is in latex:
$\sum_{k=v-1}^{n}\frac{1}{(v+1)^{k-v-1}}=\frac{1-(\frac{1}{v+1})^{n-v}}{1-\frac{1}{v+1}}$

LaTeX fix
• Jun 2nd 2011, 04:29 PM
Plato
There is a sign error in the LaTeX above.
$\sum_{k=v+1}^{n}\frac{1}{(v+1)^{k-v-1}}=\frac{1-(\frac{1}{v+1})^{n-v}}{1-\frac{1}{v+1}}$ NOTE the $k=v+1$.

Here is what you must know: $\sum\limits_{k = A}^B {r^k } = \frac{{r^A - r^{B + 1} }}{{1 - r}},\,A < B$

Notice that: $\sum\limits_{k = v + 1}^n {\left[ {\frac{1}{{v + 1}}} \right]^{k - v - 1} } = \left[ {\frac{1}{{v + 1}}} \right]^{ - v - 1} \sum\limits_{k = v +1}^n{\left[{\frac{1}{{v + 1}}} \right]^k }$

So $r=\left[ {\frac{1}{{v + 1}}} \right],~A=v+1,~\&~B=n$

Now you carry out the substitutions.
• Jun 3rd 2011, 05:21 AM
Melsi
Solved
Thank you very much, it was the piece of information that I was missing! Glad to have it solved, I am applying the solution right away. Thank you for the latex help too!

Bye!