Abbigail Robles

2022-04-24

Per the title of this question, how does one go about calculating

$\underset{n\to \mathrm{\infty}}{lim}\frac{1}{{n}^{2}}(\mathrm{ln}\left(\frac{{2}^{n}}{{3}^{n}}\right)+\mathrm{ln}\left(\frac{{5}^{n}}{{4}^{n}}\right)+\cdots +\mathrm{ln}\left(\frac{{(3n-1)}^{n}}{{(n+2)}^{n}}\right))$

Brenton Steele

Beginner2022-04-25Added 18 answers

The k-th term is

$n\mathrm{log}\frac{(3k-1)}{k+2}=n\mathrm{log}3+\frac{5n}{3k}+\mathcal{O}\left(\frac{n}{{k}^{2}}\right)$

and so your sum is

$\frac{1}{{n}^{2}}\sum _{k=1}^{n}(n\mathrm{log}3+\frac{5n}{3k}+\mathcal{O}\left(\frac{n}{{k}^{2}}\right))$

$=\frac{1}{{n}^{2}}({n}^{2}\mathrm{log}3+\frac{5n\mathrm{log}n}{3}+\mathcal{O}\left(n\right))$

Here we used the estimate$\sum _{k=1}^{n}\frac{1}{k}=\mathrm{log}n+\mathcal{O}\left(1\right)$

Thus the sum is

$=\mathrm{log}3+\frac{5\mathrm{log}n}{3n}+\mathcal{O}\left(\frac{1}{n}\right)$

and so your limit is

$\mathrm{log}3$

and so your sum is

Here we used the estimate

Thus the sum is

and so your limit is

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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