$\lim_{n\to \infty} \sum_{k=1}^n\frac{5n}{3n^2-2k}$ without L'Hopital rule

Question

The problem is: Prove that sequence converges and find its limit if it's general term is: $$\lim_{n\to \infty}x_n = \lim_{n\to \infty} \sum_{k=1}^n\frac{5n}{3n^2-2k}$$ I tried monotone convergence theorem: $$x_n \leqslant \frac{n \cdot 5n}{3n^2-2n}$$ and limit of it is $\frac{1}{3}$ but when I do $\frac{x_n+1}{x_n}$ I get $1$, but my answer should be bigger than $1$. Be simple, please.

Answer 1

It is easy to see $$\frac{n \cdot 5n}{3n^2-2}=\sum_{k=1}^n\frac{5n}{3n^2-2}<\sum_{k=1}^n\frac{5n}{3n^2-2k}<\sum_{k=1}^n\frac{5n}{3n^2-2n}=\frac{n \cdot 5n}{3n^2-2n},$$ and $$\lim_{n\to \infty}\frac{n \cdot 5n}{3n^2-2}=\frac{n \cdot 5n}{3n^2-2n}=\frac{5}{3}.$$ So $$\lim_{n\to \infty} \sum_{k=1}^n\frac{5n}{3n^2-2k}=\frac{5}{3}.$$

Answer 2

$\lim_{n\to \infty}x_n = \lim_{n\to \infty} \sum_{k=1}^n\frac{5n}{3n^2-2k} $

$\begin{array}\\ \sum_{k=1}^n\frac{5n}{3n^2-2k} &\gt \sum_{k=1}^n\frac{5n}{3n^2}\\ &= \frac{5n}{3n^2}\sum_{k=1}^n1\\ &= \frac{5}{3}\\ \end{array} $

$\begin{array}\\ \sum_{k=1}^n\frac{5n}{3n^2-2k} &\lt \sum_{k=1}^n\frac{5n}{3n^2-2n}\\ & =\sum_{k=1}^n\frac{5}{3n-2}\\ & =\frac{5}{3-2/n}\\ & =\frac{5}{3-2/n}-\frac53+\frac53\\ & =\frac{15-(15-10/n}{3(3-2/n)}+\frac53\\ & =\frac{10}{3(3n-2)}+\frac53\\ & \to\frac53\\ \end{array} $

Answer 3

Just added for your curiosity.

Since you received good answers, let me try to work the partial sums $$S_n= \sum_{k=1}^n\frac{5n}{3n^2-2k}=\frac 52 n\sum_{k=1}^n\frac{1}{\frac32n^2-k}$$

Using generalized harmonic numbers, $$ \sum_{k=1}^n\frac 1{a-k}=H_{-a}-H_{n-a}$$ Using the asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+\frac{1}{120 p^4}+O\left(\frac{1}{p^6}\right)$$ Using it twice with $a=\frac 32 n^2$, continuing with Taylor series and multiplying the result by $\frac 52 n$, we end with $$S_n=\frac{5}{3}+\frac{5}{9 n}+\frac{65}{81 n^2}+\frac{40}{81 n^3}+\frac{106}{243 n^4}+O\left(\frac{1}{n^5}\right)$$ which, for sure, shows the right limit but also allows a quite accurate approximation of the partial sums.

For example, $S_{10}=\frac{68014813111950755}{39297024420269616}\approx 1.73078787$ while the above truncated series would give $\frac{2102903}{1215000}\approx 1.73078436$.

Making the problem more general $$T_n= \sum_{k=1}^n\frac{\alpha n}{\beta n^2-\gamma k}$$ doing the same would lead to $$T_n=\frac{\alpha }{\beta }+\frac{\alpha \gamma }{2 \beta ^2 n}+\frac{\alpha \gamma (3 \beta +2 \gamma )}{6 \beta ^3 n^2}+\frac{\alpha \gamma ^2 (2 \beta +\gamma )}{4 \beta ^4 n^3}+\frac{\alpha \gamma ^2 \left(5 \beta ^2+15 \beta \gamma +6 \gamma ^2\right)}{30 \beta ^5 n^4}+O\left(\frac{1}{n^5}\right)$$