Chapter 11: Problem 4

Do the following: a. Compute: \(\lim _{n \rightarrow \infty} n \ln \left(1-\frac{3}{n}\right)\). b. Use L'Hopital's Rule to evaluate \(L=\lim _{x \rightarrow \infty}\left(1-\frac{4}{x}\right)^{x}\). [Hint: Consider \(\ln L\).] c. Determine the convergence of \(\sum_{n=1}^{\infty}\left(\frac{n}{3 n+2}\right)^{n^{2}}\). d. Sum the series \(\sum_{n=1}^{\infty}\left[\tan ^{-1} n-\tan ^{-1}(n+1)\right]\) by first writing the \(N\) th partial sum and then computing \(\lim _{N \rightarrow \infty} s_{N}\).

Short Answer

Expert verified

a. The limit is -3. b. The limit of L is \(e^{-4}\). c. The series diverges. d. The sum of the series is \(\pi/4\).

Step by step solution

Solve Problem a

Recognising the limit is in the indeterminate form \(-\infty \cdot 0\), rewrite the limit as a quotient of two functions, \( -\frac{\ln\left(1-\frac{3}{n}\right)}{1/n}\). Now, the limit appears in the `0/0` indeterminate form and can be evaluated using L'Hopital's rule. Applying L'Hopital's rule repeatedly until a determinate form emerges, the limit will be computed.

Solve Problem b

Taking the natural logarithm of L gives \(\ln(L) = x \ln \left(1-\frac{4}{x}\right)\) which is in the form `0 * infinity`. We can convert this expression into a `0/0` form by writing it as \(\frac{\ln \left(1-\frac{4}{x}\right)}{1/x}\). Now we can apply L'Hopital's rule until we get a determinate form to find the limit of L.

Solve Problem c

In order to determine the convergence of the series, the root test is used here as it is more effective given the form of the series. The root test involves taking the nth root of the absolute value of the general term and taking the limit as n goes to infinity. If this limit is less than 1, the series converges; if it is greater than 1, it diverges; otherwise, the test is inconclusive.

Solve Problem d

We recognize that the series given is a telescoping series. The Nth partial sum of the series can be written by adding up the first N terms. The terms will cancel out in such a way that only a few terms remain after cancellation. To find the sum of the series, compute the limit of the Nth partial sum as N goes to infinity.