Question

Inn 6. Test for convergence the sequence . n n=1

Answer 1

A sequence { a_{​​​​​​n} } is sait to converge if

$\large \lim_{n\to\ \infty} a_{n} =l$

Where $\large "l"$ is unique and finite. And $\large "l"$ is

called limit of sequence {a_{​​​​​​n} } .

We have sequence

$\large \left \{ \frac{ln \ n}{n} \right \}_{n=1}^{\infty} \\ \\ \\=> a_{n} = \frac{ln\ n}{n} \\ \\ \\ \lim_{n\to\infty} \ a_{n} = \lim_{n\to\infty} \frac{ln\ n}{n}= \frac{\infty}{\infty} \\ \\ \\ which\ is\ indeterminate\ form \ of\ limit. \\ \\ \\ so\ we\ will\ apply\ L'Hospital\ rule\ to\ evaluate\ limit- \\ \\ \\ According \ to\ L'Hospital\ rule\ if- \\ \\ \\ \lim_{n\to\ a}\frac{f(n)}{g(n)}= \frac{\infty}{\infty}, \ then\ \lim_{n\to\ a}\frac{f(n)}{g(n)}= \lim_{n\to\ a}\frac{f('n)}{g'(n)} \\ \\ where \ 'a' \ can\ be\ finite\ or\ infinite. \\ \\ \\ so\ by\ L'Hospital\ rule- \\ \\ \\ \lim_{n\to\infty} \frac{ln\ n}{n}= \lim_{n\to\infty} \frac{(ln\ n)'}{(n)'}= \lim_{n\to\infty} \frac{\frac{1}{n}}{1}= \lim_{n\to\infty} \frac{1}{n}= 0$

=> lim an = 0 n>00 =>1= 0, which is finite and unique. => given sequence is CONVERGENT.