\sum _{n=7}^{\infty }\:\frac{4}{11\sqrt{n}-5}

asked by guest
on Nov 15, 2024 at 7:52 am



You asked:

Evaluate the expression: \(\sum_{n = 7}^{\infty} {\frac{4}{11 \cdot \sqrt{n} - 5}}\)

MathBot Answer:

The infinite series \(\displaystyle\sum_{n=7}^{\infty} \frac{4}{11 \sqrt{n} - 5}\) diverges.


We can use Raabe-Duhamel's test to determine convergence of this series.

Raabe-Duhamel's test is an extension of the ratio test which can sometimes be used in cases where the ratio test is inconclusive.
Given a series \(\displaystyle\sum a_n\), define \(\rho = \displaystyle\lim_{n \to \infty} n\left(\frac{a_n}{a_{n+1}}-1\right)\)
 •If \(\rho > 1\) the series converges.
 •If \(\rho < 1\) the series diverges.
 •If \(\rho = 1\) or the limit does not exist the test is inconclusive.
$$\begin{aligned}L &= \lim_{n \to \infty}\left|\frac{\frac{4}{11 \sqrt{n + 1} - 5}}{\frac{4}{11 \sqrt{n} - 5}}\right| \\ &= \lim_{n \to \infty} \left|{\frac{11 \sqrt{n} - 5}{11 \sqrt{n + 1} - 5}}\right| \\ &= \frac{1}{2}\end{aligned}$$Therefore the series diverges.