Mixing
For which values of $a$ does the series $sum_{i=1}^n frac{(-2a)^n}{n^2}$ converge?
I have this series: $sum_{i=1}^n frac{(-2a)^n}{n^2}$ and the question is for what values of $a$ does the series converge. I'm supposed to write the interval $a$ is in. I first tried the ratio test:
$$ frac{(-2a)^{n+1}}{(n+1)^2} cdot frac{(n)^2}{(-2a)^n} = frac{-2a cdot n^2}{(n+1)^2} = frac{-2a cdot n^2}{n^2+2n+1}.$$
I then divided everything with $n^2$:
$$ frac{-2a cdot n^2}{n^2+2n+1} = frac{-2a}{1+2/n+1/n^2},$$
and when I took the limit as $n to infty$, I ended up with the inequality:
$$ -2a < 1. $$
I divided with $-2$ on both side, and got that $a$ is $-1/2$, but I'm supposed to make an interval, what did I do wrong?
sequences-and-series convergence
edited Nov 21 '18 at 15:24
Xander Henderson
14.1k103554
asked Nov 21 '18 at 15:21
Sherya
152
add a comment |
I have this series: $sum_{i=1}^n frac{(-2a)^n}{n^2}$ and the question is for what values of $a$ does the series converge. I'm supposed to write the interval $a$ is in. I first tried the ratio test:
$$ frac{(-2a)^{n+1}}{(n+1)^2} cdot frac{(n)^2}{(-2a)^n} = frac{-2a cdot n^2}{(n+1)^2} = frac{-2a cdot n^2}{n^2+2n+1}.$$
I then divided everything with $n^2$:
$$ frac{-2a cdot n^2}{n^2+2n+1} = frac{-2a}{1+2/n+1/n^2},$$
and when I took the limit as $n to infty$, I ended up with the inequality:
$$ -2a < 1. $$
I divided with $-2$ on both side, and got that $a$ is $-1/2$, but I'm supposed to make an interval, what did I do wrong?
sequences-and-series convergence
edited Nov 21 '18 at 15:24
Xander Henderson
14.1k103554
asked Nov 21 '18 at 15:21
Sherya
152
$$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
– Xander Henderson
Nov 21 '18 at 15:26
add a comment |
I have this series: $sum_{i=1}^n frac{(-2a)^n}{n^2}$ and the question is for what values of $a$ does the series converge. I'm supposed to write the interval $a$ is in. I first tried the ratio test:
$$ frac{(-2a)^{n+1}}{(n+1)^2} cdot frac{(n)^2}{(-2a)^n} = frac{-2a cdot n^2}{(n+1)^2} = frac{-2a cdot n^2}{n^2+2n+1}.$$
I then divided everything with $n^2$:
$$ frac{-2a cdot n^2}{n^2+2n+1} = frac{-2a}{1+2/n+1/n^2},$$
and when I took the limit as $n to infty$, I ended up with the inequality:
$$ -2a < 1. $$
I divided with $-2$ on both side, and got that $a$ is $-1/2$, but I'm supposed to make an interval, what did I do wrong?
sequences-and-series convergence
edited Nov 21 '18 at 15:24
Xander Henderson
14.1k103554
asked Nov 21 '18 at 15:21
Sherya
152
I have this series: $sum_{i=1}^n frac{(-2a)^n}{n^2}$ and the question is for what values of $a$ does the series converge. I'm supposed to write the interval $a$ is in. I first tried the ratio test:
$$ frac{(-2a)^{n+1}}{(n+1)^2} cdot frac{(n)^2}{(-2a)^n} = frac{-2a cdot n^2}{(n+1)^2} = frac{-2a cdot n^2}{n^2+2n+1}.$$
I then divided everything with $n^2$:
$$ frac{-2a cdot n^2}{n^2+2n+1} = frac{-2a}{1+2/n+1/n^2},$$
and when I took the limit as $n to infty$, I ended up with the inequality:
$$ -2a < 1. $$
I divided with $-2$ on both side, and got that $a$ is $-1/2$, but I'm supposed to make an interval, what did I do wrong?
sequences-and-series convergence
sequences-and-series convergence
edited Nov 21 '18 at 15:24
Xander Henderson
14.1k103554
asked Nov 21 '18 at 15:21
Sherya
152
edited Nov 21 '18 at 15:24
Xander Henderson
14.1k103554
asked Nov 21 '18 at 15:21
Sherya
152
edited Nov 21 '18 at 15:24
Xander Henderson
14.1k103554
edited Nov 21 '18 at 15:24
Xander Henderson
14.1k103554
edited Nov 21 '18 at 15:24
Xander Henderson
14.1k103554
14.1k103554
asked Nov 21 '18 at 15:21
Sherya
152
asked Nov 21 '18 at 15:21
Sherya
152
asked Nov 21 '18 at 15:21
Sherya
152
152
$$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
– Xander Henderson
Nov 21 '18 at 15:26
add a comment |
$$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
– Xander Henderson
Nov 21 '18 at 15:26
$$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
– Xander Henderson
Nov 21 '18 at 15:26
$$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
– Xander Henderson
Nov 21 '18 at 15:26
add a comment |
2 Answers
2
active
oldest
votes
You want to take absolute values in the ratio test. The result should be that it converges if $|-2a| < 1$, i.e.
$-1/2 < a < 1/2$, and diverges if $|-2a| > 1$. But you also need to look at the case $|-2a|=1$, where the ratio test is inconclusive.
answered Nov 21 '18 at 15:27
Robert Israel
318k23208457
add a comment |
We need that
$$frac{(-2a)^n}{n^2} to 0 implies |2a|le 1$$
then
for $-1<2a<0$ the series converges by comparison test with $sum frac1{n^2}$
for $0<2a<1$ the series converges by alternating series test
answered Nov 21 '18 at 15:44
gimusi
1
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You want to take absolute values in the ratio test. The result should be that it converges if $|-2a| < 1$, i.e.
$-1/2 < a < 1/2$, and diverges if $|-2a| > 1$. But you also need to look at the case $|-2a|=1$, where the ratio test is inconclusive.
answered Nov 21 '18 at 15:27
Robert Israel
318k23208457
add a comment |
You want to take absolute values in the ratio test. The result should be that it converges if $|-2a| < 1$, i.e.
$-1/2 < a < 1/2$, and diverges if $|-2a| > 1$. But you also need to look at the case $|-2a|=1$, where the ratio test is inconclusive.
answered Nov 21 '18 at 15:27
Robert Israel
318k23208457
add a comment |
You want to take absolute values in the ratio test. The result should be that it converges if $|-2a| < 1$, i.e.
$-1/2 < a < 1/2$, and diverges if $|-2a| > 1$. But you also need to look at the case $|-2a|=1$, where the ratio test is inconclusive.
answered Nov 21 '18 at 15:27
Robert Israel
318k23208457
You want to take absolute values in the ratio test. The result should be that it converges if $|-2a| < 1$, i.e.
$-1/2 < a < 1/2$, and diverges if $|-2a| > 1$. But you also need to look at the case $|-2a|=1$, where the ratio test is inconclusive.
answered Nov 21 '18 at 15:27
Robert Israel
318k23208457
answered Nov 21 '18 at 15:27
Robert Israel
318k23208457
answered Nov 21 '18 at 15:27
Robert Israel
318k23208457
answered Nov 21 '18 at 15:27
Robert Israel
318k23208457
318k23208457
add a comment |
add a comment |
We need that
$$frac{(-2a)^n}{n^2} to 0 implies |2a|le 1$$
then
for $-1<2a<0$ the series converges by comparison test with $sum frac1{n^2}$
for $0<2a<1$ the series converges by alternating series test
answered Nov 21 '18 at 15:44
gimusi
1
add a comment |
We need that
$$frac{(-2a)^n}{n^2} to 0 implies |2a|le 1$$
then
for $-1<2a<0$ the series converges by comparison test with $sum frac1{n^2}$
for $0<2a<1$ the series converges by alternating series test
answered Nov 21 '18 at 15:44
gimusi
1
add a comment |
We need that
$$frac{(-2a)^n}{n^2} to 0 implies |2a|le 1$$
then
for $-1<2a<0$ the series converges by comparison test with $sum frac1{n^2}$
for $0<2a<1$ the series converges by alternating series test
answered Nov 21 '18 at 15:44
gimusi
1
We need that
$$frac{(-2a)^n}{n^2} to 0 implies |2a|le 1$$
then
for $-1<2a<0$ the series converges by comparison test with $sum frac1{n^2}$
for $0<2a<1$ the series converges by alternating series test
answered Nov 21 '18 at 15:44
gimusi
1
answered Nov 21 '18 at 15:44
gimusi
1
answered Nov 21 '18 at 15:44
gimusi
1
answered Nov 21 '18 at 15:44
gimusi
1
1
add a comment |
add a comment |
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$$ -2a < 1 implies a > -frac{1}{2} implies a in left(-frac{1}{2}, inftyright), $$ but this isn't the correct interval, as you should be considering the absolute value of the ratio of $a_{n+1}/a_n$, not the ratio itself.
– Xander Henderson
Nov 21 '18 at 15:26