I tried to convert it to a form like $ \displaystyle \sum_{r=0}^n\frac{r(-1^r)}{n^{({}^{n-1}C_{r-1})}} $
And then write the series backwards and add the two, but this doesn't seem to help.
Any hints would be appreciated.
Asked
Active
Viewed 40 times
0
-
Just a comment that $^nC_r$ is not universal notation, so if you use it in a question be sure to define what you mean. It context I'm guessing you mean n choose r or $\binom{n}{r}$, as I'm used to, but it took me a minute to guess this. It may help to fix your formatting in the later sum as well. – ggg Jan 11 '21 at 17:56
-
Nice! The result is $0$ for odd $n$ and $\frac{2 n+1}{n+1}$ for even $n$. – Raffaele Jan 11 '21 at 18:19
-
@Raffaele I think you mean $2(n+1)/(n+2)$ for even $n$. – Robert Israel Jan 11 '21 at 18:26
-
@RobertIsrael No, no. It's $$\left{\frac{3}{2},\frac{5}{3},\frac{7}{4},\frac{9}{5},\frac{11}{6},\ldots\right}$$ – Raffaele Jan 11 '21 at 18:38
-
@Raffaele For example, if $n=2$, $$\sum_{r=0}^2 (-1)^r/{2 \choose r} = \frac{3}{2} = \frac{2 (n+1)}{n+2}$$ You would have $$\dfrac{2n+1}{n+1}=\frac{5}{3}$$ – Robert Israel Jan 11 '21 at 21:10
-
It's the same sequence shifted by $1$. I get $3/2$ for $n=1$ and you for $n=2$ – Raffaele Jan 11 '21 at 21:40
-
@ggg Thanks for the comment, I'll keep that in mind for future questions. – Mach Jan 12 '21 at 10:36