Calculus Multiple Choice

**Nimmy** · February 23rd 2006, 03:30 PM

Mutliple Choice questions

5. Determine which converges:

a. $\sum^{\infty}_{n=1}(4+(-1)^n)$ .
b. $\sum^{\infty}_{n=1}2^n/(n+1)!$ .
c. $\sum^{\infty}_{n=0} 5(3/2)^n$ .
d. $\sum^{\infty}_{n=1} 1/n^(1/2)$ .
e. None of these

7. Determine which test can be proved the divergence $\sum^{\infty}_{n=1} 1/n^(1/2)$ .

a. Geometric Series Test
b. P Series Test
c. Ratio Test
d. Nth Term Test for Divergence
e. None of These

14. Determine which of the following test could be used to show that $\sum^{\infty}_{n=1} (2n-1)/(3n+5)$ .

a. Root Test
b. Ratio Test
c. Geometric Series Test
d. P-Series Test
e. None of These

17. Find the radius of convergence of the power series $\sum^{\infty}_{n=0} (x/2)^n$ .

a. 1/2
b. 2
c. Infinity
d. 0
e. None of These

**ThePerfectHacker** · February 23rd 2006, 05:11 PM

Problem 5:
a)This diverges by oscillation.
b)This converges.
c)This is a geometric series with the constant ration $|3/2|>1$ thus it diverges.
d)This is called the p-series, since the exponent is less than 1 it must diverge.

Explaination to b:
Apply the ratio test,
$a_{k+1}=\frac{2^{k+1}}{(k+2)!}$
$a_k=\frac{2^k}{(k+1)!}$
Thus,
$\lim_{k\to\infty}\frac{2^{k+1}}{(k+2)!}\frac{(k+1) !}{2^k}=\frac{2}{k+2}=0<1$
Thus, it converges.
Q.E.D.

**Nimmy** · February 23rd 2006, 07:52 PM

Originally Posted by Nimmy

Mutliple Choice questions

5. Determine which converges:

a. $\sum^{\infty}_{n=1}(4+(-1)^n)$ .
b. $\sum^{\infty}_{n=1}2^n/(n+1)!$ .
c. $\sum^{\infty}_{n=0} 5(3/2)^n$ .
d. $\sum^{\infty}_{n=1} 1/n^(1/2)$ .
e. None of these

7. Determine which test can be proved the divergence $\sum^{\infty}_{n=1} 1/n^(1/2)$ .

a. Geometric Series Test
b. P Series Test
c. Ratio Test
d. Nth Term Test for Divergence
e. None of These

14. Determine which of the following test could be used to show that $\sum^{\infty}_{n=1} (2n-1)/(3n+5)$ .

a. Root Test
b. Ratio Test
c. Geometric Series Test
d. P-Series Test
e. None of These

17. Find the radius of convergence of the power series $\sum^{\infty}_{n=0} (x/2)^n$ .

a. 1/2
b. 2
c. Infinity
d. 0
e. None of These

I still need help on 7,14,17.

**ThePerfectHacker** · February 24th 2006, 08:29 AM

Originally Posted by Nimmy

I still need help on 7,14,17.

7)That is the p-series, because it has form $\frac{1}{k^n}$ .

14)None of these, you may solve it using a different test. The limit comparison test.

17)Use the ratio test, $a_k=(x/2)^k$ and $a_{k+1}=(x/2)^{k+1}$ .
Thus,
$\lim_{k\to\infty}\frac{x^{k+1}}{2^{k+1}}\frac{2^k} {x^k}=\frac{x}{2}$ Now it is necessary for this is converge when $|\frac{x}{2}|<1$ that happens when $-2<x<2$ thus the radius of convergence is 2.

**Rich B.** · February 24th 2006, 01:11 PM

Hi Folks:

Are we doing homework or completing take-home tests for students of dimished ambition? ...just a thought (not an insinuation).

Respectfully,

Rich B.

Thread: Calculus Multiple Choice

LinkBack

Thread Tools

Display

Calculus Multiple Choice

Similar Math Help Forum Discussions

Multiple Choice

AP Calculus Multiple Choice Questions- Help!

Multiple Choice

Multiple choice

Calculus Problems Multiple Choice

Search Tags