hawatajwizp

2022-06-24

Find the limit of $\frac{{T}_{n}}{5n+4}$

Braedon Rivas

Assuming we know that ${U}_{n}$ converges (which is the premise of the question), we see that
$\underset{n\to \mathrm{\infty }}{lim}{U}_{n+1}=\underset{n\to \mathrm{\infty }}{lim}{U}_{n}.$
Then,
$\underset{n\to \mathrm{\infty }}{lim}\frac{{U}_{n}+3}{5-{U}_{n}}=\underset{n\to \mathrm{\infty }}{lim}{U}_{n}.$
Then, in the limit,
$\frac{{U}_{n}+3}{5-{U}_{n}}={U}_{n},$
and solving for ${U}_{n}$ gives ${U}_{n}=1$ or ${U}_{n}=3.$ However, since ${U}_{0}=0,$ it is clear that ${U}_{n}$ cannot exceed 1 and hence, $\underset{n\to \mathrm{\infty }}{lim}{U}_{n}=1$
Again, the premise of the question assumes that $\underset{n\to \mathrm{\infty }}{lim}\frac{{T}_{n}}{5n+4}$ converges. Then, we can write
$\underset{n\to \mathrm{\infty }}{lim}\frac{{T}_{n+1}}{5\left(n+1\right)+4}=\underset{n\to \mathrm{\infty }}{lim}\frac{{T}_{n}}{5n+4}.$
Further, we see that, through the sum,
${T}_{n+1}=\sum _{k=1}^{n+1}\frac{1}{{U}_{k}-3}=\left(\sum _{k=1}^{n}\frac{1}{{U}_{k}-3}\right)+\frac{1}{{U}_{n+1}-3}={T}_{n}+\frac{1}{{U}_{n+1}-3}.$
We also know that $\underset{n\to \mathrm{\infty }}{lim}{U}_{n+1}=1,$ so $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{U}_{n+1}-3}=-\frac{1}{2}.$ Thus, in the limit, we have
$\frac{{T}_{n+1}}{5\left(n+1\right)+4}=\frac{{T}_{n}-\frac{1}{2}}{5n+9}=\frac{{T}_{n}}{5n+4}.$
Finally, solving for ${T}_{n}$ gives
${T}_{n}=-\frac{1}{10}\left(5n+4\right),$
so
$\underset{n\to \mathrm{\infty }}{lim}\frac{{T}_{n}}{5n+4}=-\frac{1}{10}.$

Do you have a similar question?