What is the value of the convergent series $ sum_n=1^infty (e^-4n-e^-4(n-1))$? The Next CEO of Stack OverflowConvergent/divergent series $sum_n=2^infty(nsqrtn-sqrtn^3-1)$Does $sum_n = 1^inftyfracln^2(n)sqrtn(8n + 9sqrtn)$ Converge?Find values of $p$ for which the series is convergent.The Series( $sum_1^+ inftyfrac1n! + n$ convergence or divergence?$sum_n= 0^inftya_n$ converges, what other series must then also converge?Trying to study to convergence of the series $sum_n=1^infty frac (n!)^2 4^n (2n)!$For what P would the series $sum_n=2^infty frac1n^pln(n)$ converge?Determining whether the series: $sum_n=1^infty tanleft(frac1nright) $ converges$sum_limitsn=1^infty(1-cos(fracpin))$ convergence proofConvergence of the series $sum_n=1^infty a^ln(n)$
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What is the value of the convergent series $ sum_n=1^infty (e^-4n-e^-4(n-1))$?
The Next CEO of Stack OverflowConvergent/divergent series $sum_n=2^infty(nsqrtn-sqrtn^3-1)$Does $sum_n = 1^inftyfracln^2(n)sqrtn(8n + 9sqrtn)$ Converge?Find values of $p$ for which the series is convergent.The Series( $sum_1^+ inftyfrac1n! + n$ convergence or divergence?$sum_n= 0^inftya_n$ converges, what other series must then also converge?Trying to study to convergence of the series $sum_n=1^infty frac (n!)^2 4^n (2n)!$For what P would the series $sum_n=2^infty frac1n^pln(n)$ converge?Determining whether the series: $sum_n=1^infty tanleft(frac1nright) $ converges$sum_limitsn=1^infty(1-cos(fracpin))$ convergence proofConvergence of the series $sum_n=1^infty a^ln(n)$
$begingroup$
Consider the series
$$
sum_n=1^infty (e^-4n-e^-4(n-1)).
$$
I know the series must converge but, what test should I employ to find the value of the series?
Attempt:
I have tried using the integral test, which gives a value of 3/2, but it is incorrect.
The following is the original problem:
calculus sequences-and-series limits
edited Mar 19 at 16:21
Jack
27.6k1782203
asked Mar 16 at 18:55
MCCMCC
316
$endgroup$
add a comment |
$begingroup$
Consider the series
$$
sum_n=1^infty (e^-4n-e^-4(n-1)).
$$
I know the series must converge but, what test should I employ to find the value of the series?
Attempt:
I have tried using the integral test, which gives a value of 3/2, but it is incorrect.
The following is the original problem:
calculus sequences-and-series limits
edited Mar 19 at 16:21
Jack
27.6k1782203
asked Mar 16 at 18:55
MCCMCC
316
$endgroup$
1
$begingroup$
This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^-4n$ and you get a standard geometric sum.
$endgroup$
– Michael
Mar 16 at 19:02
$begingroup$
I assume you know how to evaluate $Asum_n=1^infty r^n$.
$endgroup$
– Michael
Mar 16 at 19:16
$begingroup$
Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
$endgroup$
– Wojowu
Mar 16 at 19:39
add a comment |
$begingroup$
Consider the series
$$
sum_n=1^infty (e^-4n-e^-4(n-1)).
$$
I know the series must converge but, what test should I employ to find the value of the series?
Attempt:
I have tried using the integral test, which gives a value of 3/2, but it is incorrect.
The following is the original problem:
calculus sequences-and-series limits
edited Mar 19 at 16:21
Jack
27.6k1782203
asked Mar 16 at 18:55
MCCMCC
316
$endgroup$
Consider the series
$$
sum_n=1^infty (e^-4n-e^-4(n-1)).
$$
I know the series must converge but, what test should I employ to find the value of the series?
Attempt:
I have tried using the integral test, which gives a value of 3/2, but it is incorrect.
The following is the original problem:
calculus sequences-and-series limits
calculus sequences-and-series limits
edited Mar 19 at 16:21
Jack
27.6k1782203
asked Mar 16 at 18:55
MCCMCC
316
edited Mar 19 at 16:21
Jack
27.6k1782203
asked Mar 16 at 18:55
MCCMCC
316
edited Mar 19 at 16:21
Jack
27.6k1782203
edited Mar 19 at 16:21
Jack
27.6k1782203
edited Mar 19 at 16:21
Jack
27.6k1782203
27.6k1782203
asked Mar 16 at 18:55
MCCMCC
316
asked Mar 16 at 18:55
MCCMCC
316
asked Mar 16 at 18:55
MCCMCC
316
316
1
$begingroup$
This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^-4n$ and you get a standard geometric sum.
$endgroup$
– Michael
Mar 16 at 19:02
$begingroup$
I assume you know how to evaluate $Asum_n=1^infty r^n$.
$endgroup$
– Michael
Mar 16 at 19:16
$begingroup$
Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
$endgroup$
– Wojowu
Mar 16 at 19:39
add a comment |
1
$begingroup$
This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^-4n$ and you get a standard geometric sum.
$endgroup$
– Michael
Mar 16 at 19:02
$begingroup$
I assume you know how to evaluate $Asum_n=1^infty r^n$.
$endgroup$
– Michael
Mar 16 at 19:16
$begingroup$
Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
$endgroup$
– Wojowu
Mar 16 at 19:39
1
1
$begingroup$
This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^-4n$ and you get a standard geometric sum.
$endgroup$
– Michael
Mar 16 at 19:02
$begingroup$
This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^-4n$ and you get a standard geometric sum.
$endgroup$
– Michael
Mar 16 at 19:02
$begingroup$
I assume you know how to evaluate $Asum_n=1^infty r^n$.
$endgroup$
– Michael
Mar 16 at 19:16
$begingroup$
I assume you know how to evaluate $Asum_n=1^infty r^n$.
$endgroup$
– Michael
Mar 16 at 19:16
$begingroup$
Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
$endgroup$
– Wojowu
Mar 16 at 19:39
$begingroup$
Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
$endgroup$
– Wojowu
Mar 16 at 19:39
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
This series converges. Here is the reason:$$sum_n=1^infty e^-4n-e^-4(n+1)=sum_n=1^infty e^-4n-e^-4ne^-4\=(1-e^-4)sum_n=1^inftye^-4n\=(1-e^-4)e^-4over 1-e^-4\=e^-4$$
answered Mar 16 at 19:31
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
$begingroup$
The partial sums are $e^-4-e^-4(n+1)$ by induction, making the limit $e^-4$.
answered Mar 16 at 19:40
J.G.J.G.
32.5k23250
$endgroup$
3
$begingroup$
Exploiting the telescoping series is a much more expedient way forward.
$endgroup$
– Mark Viola
Mar 16 at 20:54
add a comment |
$begingroup$
Split the sum$$beginalign*sumlimits_ngeq1left[e^-4n-e^-4(n-1)right] & =sumlimits_ngeq1e^-4n-sumlimits_ngeq1e^-4(n-1)\ & =e^-4sumlimits_ngeq0e^-4n-sumlimits_ngeq0e^-4n\ & =left(e^-4-1right)sumlimits_ngeq0e^-4nendalign*$$The infinite sum converges by the infinite geometric sequence which converges for $|x|<1$.$$sumlimits_ngeq0x^n=frac 11-x$$It can be shown that $e^-4<1$, meaning the sum converges.
answered Mar 16 at 19:40
Frank W.Frank W.
3,7351321
$endgroup$
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This series converges. Here is the reason:$$sum_n=1^infty e^-4n-e^-4(n+1)=sum_n=1^infty e^-4n-e^-4ne^-4\=(1-e^-4)sum_n=1^inftye^-4n\=(1-e^-4)e^-4over 1-e^-4\=e^-4$$
answered Mar 16 at 19:31
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
$begingroup$
This series converges. Here is the reason:$$sum_n=1^infty e^-4n-e^-4(n+1)=sum_n=1^infty e^-4n-e^-4ne^-4\=(1-e^-4)sum_n=1^inftye^-4n\=(1-e^-4)e^-4over 1-e^-4\=e^-4$$
answered Mar 16 at 19:31
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
$begingroup$
This series converges. Here is the reason:$$sum_n=1^infty e^-4n-e^-4(n+1)=sum_n=1^infty e^-4n-e^-4ne^-4\=(1-e^-4)sum_n=1^inftye^-4n\=(1-e^-4)e^-4over 1-e^-4\=e^-4$$
answered Mar 16 at 19:31
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
This series converges. Here is the reason:$$sum_n=1^infty e^-4n-e^-4(n+1)=sum_n=1^infty e^-4n-e^-4ne^-4\=(1-e^-4)sum_n=1^inftye^-4n\=(1-e^-4)e^-4over 1-e^-4\=e^-4$$
answered Mar 16 at 19:31
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 16 at 19:31
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 16 at 19:31
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 16 at 19:31
Mostafa AyazMostafa Ayaz
18.3k31040
18.3k31040
add a comment |
add a comment |
$begingroup$
The partial sums are $e^-4-e^-4(n+1)$ by induction, making the limit $e^-4$.
answered Mar 16 at 19:40
J.G.J.G.
32.5k23250
$endgroup$
3
$begingroup$
Exploiting the telescoping series is a much more expedient way forward.
$endgroup$
– Mark Viola
Mar 16 at 20:54
add a comment |
$begingroup$
The partial sums are $e^-4-e^-4(n+1)$ by induction, making the limit $e^-4$.
answered Mar 16 at 19:40
J.G.J.G.
32.5k23250
$endgroup$
3
$begingroup$
Exploiting the telescoping series is a much more expedient way forward.
$endgroup$
– Mark Viola
Mar 16 at 20:54
add a comment |
$begingroup$
The partial sums are $e^-4-e^-4(n+1)$ by induction, making the limit $e^-4$.
answered Mar 16 at 19:40
J.G.J.G.
32.5k23250
$endgroup$
The partial sums are $e^-4-e^-4(n+1)$ by induction, making the limit $e^-4$.
answered Mar 16 at 19:40
J.G.J.G.
32.5k23250
answered Mar 16 at 19:40
J.G.J.G.
32.5k23250
answered Mar 16 at 19:40
J.G.J.G.
32.5k23250
answered Mar 16 at 19:40
J.G.J.G.
32.5k23250
32.5k23250
3
$begingroup$
Exploiting the telescoping series is a much more expedient way forward.
$endgroup$
– Mark Viola
Mar 16 at 20:54
add a comment |
3
$begingroup$
Exploiting the telescoping series is a much more expedient way forward.
$endgroup$
– Mark Viola
Mar 16 at 20:54
3
3
$begingroup$
Exploiting the telescoping series is a much more expedient way forward.
$endgroup$
– Mark Viola
Mar 16 at 20:54
$begingroup$
Exploiting the telescoping series is a much more expedient way forward.
$endgroup$
– Mark Viola
Mar 16 at 20:54
add a comment |
$begingroup$
Split the sum$$beginalign*sumlimits_ngeq1left[e^-4n-e^-4(n-1)right] & =sumlimits_ngeq1e^-4n-sumlimits_ngeq1e^-4(n-1)\ & =e^-4sumlimits_ngeq0e^-4n-sumlimits_ngeq0e^-4n\ & =left(e^-4-1right)sumlimits_ngeq0e^-4nendalign*$$The infinite sum converges by the infinite geometric sequence which converges for $|x|<1$.$$sumlimits_ngeq0x^n=frac 11-x$$It can be shown that $e^-4<1$, meaning the sum converges.
answered Mar 16 at 19:40
Frank W.Frank W.
3,7351321
$endgroup$
add a comment |
$begingroup$
Split the sum$$beginalign*sumlimits_ngeq1left[e^-4n-e^-4(n-1)right] & =sumlimits_ngeq1e^-4n-sumlimits_ngeq1e^-4(n-1)\ & =e^-4sumlimits_ngeq0e^-4n-sumlimits_ngeq0e^-4n\ & =left(e^-4-1right)sumlimits_ngeq0e^-4nendalign*$$The infinite sum converges by the infinite geometric sequence which converges for $|x|<1$.$$sumlimits_ngeq0x^n=frac 11-x$$It can be shown that $e^-4<1$, meaning the sum converges.
answered Mar 16 at 19:40
Frank W.Frank W.
3,7351321
$endgroup$
add a comment |
$begingroup$
Split the sum$$beginalign*sumlimits_ngeq1left[e^-4n-e^-4(n-1)right] & =sumlimits_ngeq1e^-4n-sumlimits_ngeq1e^-4(n-1)\ & =e^-4sumlimits_ngeq0e^-4n-sumlimits_ngeq0e^-4n\ & =left(e^-4-1right)sumlimits_ngeq0e^-4nendalign*$$The infinite sum converges by the infinite geometric sequence which converges for $|x|<1$.$$sumlimits_ngeq0x^n=frac 11-x$$It can be shown that $e^-4<1$, meaning the sum converges.
answered Mar 16 at 19:40
Frank W.Frank W.
3,7351321
$endgroup$
Split the sum$$beginalign*sumlimits_ngeq1left[e^-4n-e^-4(n-1)right] & =sumlimits_ngeq1e^-4n-sumlimits_ngeq1e^-4(n-1)\ & =e^-4sumlimits_ngeq0e^-4n-sumlimits_ngeq0e^-4n\ & =left(e^-4-1right)sumlimits_ngeq0e^-4nendalign*$$The infinite sum converges by the infinite geometric sequence which converges for $|x|<1$.$$sumlimits_ngeq0x^n=frac 11-x$$It can be shown that $e^-4<1$, meaning the sum converges.
answered Mar 16 at 19:40
Frank W.Frank W.
3,7351321
answered Mar 16 at 19:40
Frank W.Frank W.
3,7351321
answered Mar 16 at 19:40
Frank W.Frank W.
3,7351321
answered Mar 16 at 19:40
Frank W.Frank W.
3,7351321
3,7351321
add a comment |
add a comment |
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$begingroup$
This is a telescoping series. Try writing out the first few terms to see the cancellations. Alternatively you can just factor out $e^-4n$ and you get a standard geometric sum.
$endgroup$
– Michael
Mar 16 at 19:02
$begingroup$
I assume you know how to evaluate $Asum_n=1^infty r^n$.
$endgroup$
– Michael
Mar 16 at 19:16
$begingroup$
Assuming the terms of the series are monotonic, the only thing you can get out of the integral test is whether the sum converges. You cannot compute the value of the series this way.
$endgroup$
– Wojowu
Mar 16 at 19:39