JEE limits question

$ln(y) = \dfrac{1}{n} \lim_{n\to\infty} \left(\sum_{p=1}^{n}\sum_{k=1}^{p-1} \int_{k}^{k+1} \dfrac{k+1}{x(x+1)}dx - n\ln(n+1)\right)$

Let us break the problem into pieces. First let us solve: $\sum_{k=1}^{p-1} \int_{k}^{k+1} \dfrac{k+1}{x(x+1)}dx$

Upon solving the definite integral we get:

$\sum_{k=1}^{p-1} (k+1)(2\ln(k+1)-\ln(k+2)-lnk)$

By rearranging the abouve expression it is obvious to notice the difference series here:

$\left(\sum_{k=1}^{p-1} (k+1)\ln(k+1)-klnk\right)-\left(\sum_{k=1}^{p-1} (k+1)\ln(k+2)-k\ln(k+1)\right)+\left(\sum_{k=1}^{p-1} \ln(k+1)-lnk\right)$

Upon simplifying we get:

$\left(plnp\right)+\left(-p\ln(p+1)+ln2\right)+\left(lnp\right) = \ln\left(\dfrac{2p^{(p+1)}}{(p+1)^p}\right)$

Now moving towards the second part of the problem:

$\sum_{p=1}^{n} \ln\left(\dfrac{2p^{(p+1)}}{(p+1)^p}\right)- n\ln(n+1)$

FACT : We know the sum of the logarithm is the product of its argument:

$\ln\left(2^n\dfrac{1^2 2^3 3^4 \cdots n^{n+1}}{2^1 3^2 4^3 \cdots n^{n-1} (n+1)^n}\right)- n\ln(n+1)$ $= \ln\left(2^n\dfrac{(n!)^2}{(n+1)^n}\right)- n\ln(n+1)$ $= \ln\left(2^n\dfrac{(n!)^2}{(n+1)^{2n}}\right)$

Now applying the limit

$\dfrac{1}{n} \lim_{n\to\infty} \ln\left(2^n\left(\dfrac{(n!)}{(n+1)^n}\right)^2\right)$

Again using the FACT we convert the above expression to a sum

$\dfrac{1}{n} \lim_{n\to\infty}\sum_{r=1}^{n} \ln\left(\left(2\dfrac{(r)}{(n+1)}\right)^2\right)$

Clearly above the second term cancels to zero.

$= \dfrac{1}{n} \lim_{n\to\infty}\sum_{r=1}^{n} \ln\left(\left(2\dfrac{(r)}{(n+1)}\right)^2\right)$

The Reimann Sum can be used here as well as n+1 is essentially equal to n

$= \int_{0}^{1} \ln(2t^2)$

$ln(y) = ln2-2$

$\ln(\dfrac{y}{2})+3=1$