# Thread: Improper Integral, Sequences, and Series

1. ## Improper Integral, Sequences, and Series

Hey guys, I have a few problems that if possible, I would love to have someone check my solutions to them.

Problem 1. Determine whether the improper integral $\displaystyle \int^{\infty}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx$ converges, or diverges. In the case of convergence, give its value.

My Solution

$\displaystyle = \lim_{t \to \infty}\int^{t}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx = \lim_{t \to \infty}[lnx - ln(x^2 + 1)]^{t}_{1} = \lim_{t \to \infty}[(lnt - ln(t^2 + 1) - (ln1 - ln2)] =$ $\displaystyle \lim_{t \to \infty}(lnt - ln(t^2 + 1) + ln2) = \infty - \infty + ln2 = ln2$ ... Converges to ln2

Problem 2.
Determine whether the sequence with the given general term converges, or diverges. In the case of convergence, give its value.

a.) $\displaystyle a_{n} = \frac{2n - 1}{3 - 5n}$

My Solution
$\displaystyle \lim_{n \to \infty}\frac{2n - 1}{3 - 5n} = \frac{\infty}{\infty}$, Using L'Hopital's Rule. $\displaystyle = \lim_{n \to \infty}\frac{-2}{5} = \frac{-2}{5}$ ... Converges to $\displaystyle \frac{-2}{5}$

b.) $\displaystyle a_{n} = (-1)^{n}\sin{\frac{n\pi}{2}}$

My Solution

$\displaystyle \lim_{n \to \infty}|(-1)^{n}\sin{\frac{n\pi}{2}}| = \lim_{n \to \infty}|(-1)^{n}|*|\sin{\frac{n\pi}{2}}| = \lim_{n \to \infty}1 * \sin{\frac{n\pi}{2}} =$ No Limit, Diverges

c.) $\displaystyle a_{n} = (1 + \frac{1}{n})^n$

My Solution

$\displaystyle 1^{\infty}$ ... $\displaystyle y = (1 + \frac{1}{n})^{n}, lny = nln(1 + \frac{1}{n})$

$\displaystyle \lim_{n \to \infty} lny = \lim_{n \to \infty}\frac{ln(1 + \frac{1}{n})}{\frac{1}{n}} = \frac{0}{0},$ Using L'Hopital's Rule.$\displaystyle = \lim_{n \to \infty}\frac{\frac{-1}{n^2 + n}}{\frac{-1}{n^2}} = \lim_{n \to \infty}\frac{n}{n + 1} = \frac{\infty}{\infty} = \lim_{n \to \infty}\frac{1}{1} = 1$ ... $\displaystyle lny \to 1,$ $\displaystyle y \to e$ ... Converges to e

Problem 3. Calculate the sum of the given convergent geometric series.

a.) $\displaystyle \frac{9}{4} - \frac{3}{2} + 1$ ...

My Solution

$\displaystyle a = \frac{9}{4}$, $\displaystyle r = \frac{-2}{3}$

$\displaystyle \frac{\frac{9}{4}}{1 + \frac{2}{3}} = \frac{27}{20}$

b.)
$\displaystyle \Sigma^{\infty}_{n = 1}(-1)^{n}(\frac{3}{7})^n$

My Solution

$\displaystyle a = \frac{-3}{7}$, $\displaystyle r = \frac{-3}{7}$

$\displaystyle \frac{\frac{-3}{7}}{1 + \frac{3}{7}} = \frac{-3}{10}$

Problem 4. Write the first three terms of the sequence of partial sums of the series $\displaystyle \Sigma^{\infty}_{k = 1}\frac{k}{2k - 1}$.

My Solution

$\displaystyle S_{1} = \frac{1}{1} = 1$
$\displaystyle S_{2} = 1 + \frac{2}{3} = \frac{5}{3}$
$\displaystyle S_{3} = \frac{5}{3} + \frac{3}{5} = \frac{34}{15}$

$\displaystyle 1$, $\displaystyle \frac{5}{3}$, $\displaystyle \frac{34}{15}$

Thank you in advance for any help!

2. Originally Posted by mturner07
Hey guys, I have a few problems that if possible, I would love to have someone check my solutions to them.

Problem 1. Determine whether the improper integral $\displaystyle \int^{\infty}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx$ converges, or diverges. In the case of convergence, give its value.

My Solution

$\displaystyle = \lim_{t \to \infty}\int^{t}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx = \lim_{t \to \infty}[lnx - ln(x^2 + 1)]^{t}_{1} = \lim_{t \to \infty}[(lnt - ln(t^2 + 1) - (ln1 - ln2)] =$ $\displaystyle \lim_{t \to \infty}(lnt - ln(t^2 + 1) + ln2) = \infty - \infty + ln2 = ln2$ ... Converges to ln2

Nop: $\displaystyle \ln t-\ln(t^2+1)=\ln\left(\frac{t}{t^2+1}\right)\xrighta rrow[t\to\infty]{}-\infty$ and thus the integral diverges.

Of course, you can't write $\displaystyle \infty-\infty=0$ ...

Problem 2. Determine whether the sequence with the given general term converges, or diverges. In the case of convergence, give its value.

a.) $\displaystyle a_{n} = \frac{2n - 1}{3 - 5n}$

My Solution
$\displaystyle \lim_{n \to \infty}\frac{2n - 1}{3 - 5n} = \frac{\infty}{\infty}$, Using L'Hopital's Rule. $\displaystyle = \lim_{n \to \infty}\frac{-2}{5} = \frac{-2}{5}$ ... Converges to $\displaystyle \frac{-2}{5}$

Nop. You can't use DIRECTLY L'Hospital with a discrete variable since L'H implies the use of derivative which use limits which need a continuous variable. Of course, you can use L'H with $\displaystyle \frac{2x-1}{3-5x}$ and then use that the limit stays the same no matter how $\displaystyle x\rightarrow \infty$ , and thus this is so if you choose to go to the limit along the naturals.

Another way, perhaps more natural , to divide both numerator and denominator by the highest power of n and use arithmetic of limits:

$\displaystyle \frac{2n-1}{3-5n}\cdot\frac{1\slash n}{1\slash n} = \frac{2-\frac{1}{n}}{\frac{3}{n}-5}\xrightarrow [n\to\infty]{}\frac{2-0}{0-5}=-\frac{2}{5}$

Tonio

b.) $\displaystyle a_{n} = (-1)^{n}\sin{\frac{n\pi}{2}}$

My Solution

$\displaystyle \lim_{n \to \infty}|(-1)^{n}\sin{\frac{n\pi}{2}}| = \lim_{n \to \infty}|(-1)^{n}|*|\sin{\frac{n\pi}{2}}| = \lim_{n \to \infty}1 * \sin{\frac{n\pi}{2}} =$ No Limit, Diverges

c.) $\displaystyle a_{n} = (1 + \frac{1}{n})^n$

My Solution

$\displaystyle 1^{\infty}$ ... $\displaystyle y = (1 + \frac{1}{n})^{n}, lny = nln(1 + \frac{1}{n})$

$\displaystyle \lim_{n \to \infty} lny = \lim_{n \to \infty}\frac{ln(1 + \frac{1}{n})}{\frac{1}{n}} = \frac{0}{0},$ Using L'Hopital's Rule.$\displaystyle = \lim_{n \to \infty}\frac{\frac{-1}{n^2 + n}}{\frac{-1}{n^2}} = \lim_{n \to \infty}\frac{n}{n + 1} = \frac{\infty}{\infty} = \lim_{n \to \infty}\frac{1}{1} = 1$ ... $\displaystyle lny \to 1,$ $\displaystyle y \to e$ ... Converges to e

Problem 3. Calculate the sum of the given convergent geometric series.

a.) $\displaystyle \frac{9}{4} - \frac{3}{2} + 1$ ...

My Solution

$\displaystyle a = \frac{9}{4}$, $\displaystyle r = \frac{-2}{3}$

$\displaystyle \frac{\frac{9}{4}}{1 + \frac{2}{3}} = \frac{27}{20}$

b.) $\displaystyle \Sigma^{\infty}_{n = 1}(-1)^{n}(\frac{3}{7})^n$

My Solution

$\displaystyle a = \frac{-3}{7}$, $\displaystyle r = \frac{-3}{7}$

$\displaystyle \frac{\frac{-3}{7}}{1 + \frac{3}{7}} = \frac{-3}{10}$

Problem 4. Write the first three terms of the sequence of partial sums of the series $\displaystyle \Sigma^{\infty}_{k = 1}\frac{k}{2k - 1}$.

My Solution

$\displaystyle S_{1} = \frac{1}{1} = 1$
$\displaystyle S_{2} = 1 + \frac{2}{3} = \frac{5}{3}$
$\displaystyle S_{3} = \frac{5}{3} + \frac{3}{5} = \frac{34}{15}$

$\displaystyle 1$, $\displaystyle \frac{5}{3}$, $\displaystyle \frac{34}{15}$

Thank you in advance for any help!
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