Question

If $f\left(n\right)=\sum _{r=1}^n\frac{1}{(\sqrt{r}+\sqrt{r+1})(r^{1/4}+(r+1{)}^{1/4})}$ and $g\left(n\right)$ is product of $n$ terms of the G.P. $16,4,1\dots ,$ then which of the following is correct?

• A. $f\left(80\right)$ is a prime number
• B. $f\left(9999\right)=10$
• C. $\sum _{n=1}^{\infty }\left(g\right(n){)}^{1/n}=32$
• D. $\underset{n\to \infty }{lim}\left(g\right(n){)}^{1/n^2}=\frac{1}{32}$

Answer 1

Answer: (A), (C)

$\sum _{n=1}^{\infty }\left(g\right(n){)}^{1/n}=32$

$\frac{1}{(\sqrt{r}+\sqrt{r+1})(r^{1/4}+(r+1{)}^{1/4})}=\frac{(\sqrt{r+1}-\sqrt{r})}{(r^{1/4}+(r+1{)}^{1/4})}=\frac{(\sqrt{r+1}-\sqrt{r})(r^{1/4}-(r+1{)}^{1/4})}{(r^{1/2}-(r+1{)}^{1/2})}=(r+1{)}^{1/4}-r^{1/4}$
Therefore,
$f\left(n\right)=\sum _{r=1}^n\left[\right(r+1{)}^{1/4}-r^{1/4}]\Rightarrow f(n)=(n+1{)}^{1/4}-1$
So,
$f\left(80\right)=(81{)}^{1/4}-1=2$
Which is a prime number.
$f\left(9999\right)=(10000{)}^{1/4}-1=9$

Now,
$g\left(n\right)=16\cdot 4\cdot 1\cdots n\text{terms}\Rightarrow g\left(n\right)=4^{2+1+0+\cdots }$
We know that
$2+1+0+\cdots n\text{terms}=\frac{n}{2}[4+(n-1)\times (-1\left)\right]=\frac{n(5-n)}{2}$
Therefore,
$g\left(n\right)=2^{n(5-n)}\Rightarrow \sum _{n=1}^{\infty }\left(g\right(n){)}^{1/n}=\sum _{n=1}^{\infty }{2}^{(5-n)}\Rightarrow \sum _{n=1}^{\infty }(g\left(n\right){)}^{1/n}=2^4+2^3+2^2+\cdots \therefore \sum _{n=1}^{\infty }\left(g\right(n){)}^{1/n}=\frac{16}{1-\frac{1}{2}}=32$

$\underset{n\to \infty }{lim}\left(g\right(n){)}^{1/n^2}=\underset{n\to \infty }{lim}{2}^{(5-n)/n}=\frac{1}{2}$