Infinite Series arising in Laplace transform
$begingroup$
I'm trying to understand how the answer was computed for the following infinite series:
$sum _ { x = 1 } ^ { infty } e ^ { - s x } p q ^ { x - 1 } = frac { p e ^ { - s } } { 1 - q e ^ { - s } }$
Any help would be highly appreciated.
sequences-and-series convergence power-series laplace-transform laurent-series
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add a comment |
$begingroup$
I'm trying to understand how the answer was computed for the following infinite series:
$sum _ { x = 1 } ^ { infty } e ^ { - s x } p q ^ { x - 1 } = frac { p e ^ { - s } } { 1 - q e ^ { - s } }$
Any help would be highly appreciated.
sequences-and-series convergence power-series laplace-transform laurent-series
$endgroup$
add a comment |
$begingroup$
I'm trying to understand how the answer was computed for the following infinite series:
$sum _ { x = 1 } ^ { infty } e ^ { - s x } p q ^ { x - 1 } = frac { p e ^ { - s } } { 1 - q e ^ { - s } }$
Any help would be highly appreciated.
sequences-and-series convergence power-series laplace-transform laurent-series
$endgroup$
I'm trying to understand how the answer was computed for the following infinite series:
$sum _ { x = 1 } ^ { infty } e ^ { - s x } p q ^ { x - 1 } = frac { p e ^ { - s } } { 1 - q e ^ { - s } }$
Any help would be highly appreciated.
sequences-and-series convergence power-series laplace-transform laurent-series
sequences-and-series convergence power-series laplace-transform laurent-series
asked Jan 28 at 18:57
AlexAlex
115
115
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1 Answer
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Hint: $e^{-sx}pq^{x-1}=(e^{-s}q)^{x-1} cdot pe^{-s}$. You know the sum of a geometric series, right?
$endgroup$
$begingroup$
Thanks for the hint @Mindlack. I know very little about geometric series (and series in general) though. I guess this now should look like some expression for which the sum is well known. I will try to compute this using online solver (although it could not give me an answer for the original form).
$endgroup$
– Alex
Jan 28 at 19:27
$begingroup$
I got it now :), many thanks @Mindlack. Just needed to replace accordingly in $S _ { infty } = sum _ { n = 1 } ^ { infty } a r ^ { n - 1 } = frac { a _ { } } { 1 - r ^ { } }$
$endgroup$
– Alex
Jan 28 at 19:35
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $e^{-sx}pq^{x-1}=(e^{-s}q)^{x-1} cdot pe^{-s}$. You know the sum of a geometric series, right?
$endgroup$
$begingroup$
Thanks for the hint @Mindlack. I know very little about geometric series (and series in general) though. I guess this now should look like some expression for which the sum is well known. I will try to compute this using online solver (although it could not give me an answer for the original form).
$endgroup$
– Alex
Jan 28 at 19:27
$begingroup$
I got it now :), many thanks @Mindlack. Just needed to replace accordingly in $S _ { infty } = sum _ { n = 1 } ^ { infty } a r ^ { n - 1 } = frac { a _ { } } { 1 - r ^ { } }$
$endgroup$
– Alex
Jan 28 at 19:35
add a comment |
$begingroup$
Hint: $e^{-sx}pq^{x-1}=(e^{-s}q)^{x-1} cdot pe^{-s}$. You know the sum of a geometric series, right?
$endgroup$
$begingroup$
Thanks for the hint @Mindlack. I know very little about geometric series (and series in general) though. I guess this now should look like some expression for which the sum is well known. I will try to compute this using online solver (although it could not give me an answer for the original form).
$endgroup$
– Alex
Jan 28 at 19:27
$begingroup$
I got it now :), many thanks @Mindlack. Just needed to replace accordingly in $S _ { infty } = sum _ { n = 1 } ^ { infty } a r ^ { n - 1 } = frac { a _ { } } { 1 - r ^ { } }$
$endgroup$
– Alex
Jan 28 at 19:35
add a comment |
$begingroup$
Hint: $e^{-sx}pq^{x-1}=(e^{-s}q)^{x-1} cdot pe^{-s}$. You know the sum of a geometric series, right?
$endgroup$
Hint: $e^{-sx}pq^{x-1}=(e^{-s}q)^{x-1} cdot pe^{-s}$. You know the sum of a geometric series, right?
edited Jan 28 at 19:14
answered Jan 28 at 19:05
MindlackMindlack
4,910211
4,910211
$begingroup$
Thanks for the hint @Mindlack. I know very little about geometric series (and series in general) though. I guess this now should look like some expression for which the sum is well known. I will try to compute this using online solver (although it could not give me an answer for the original form).
$endgroup$
– Alex
Jan 28 at 19:27
$begingroup$
I got it now :), many thanks @Mindlack. Just needed to replace accordingly in $S _ { infty } = sum _ { n = 1 } ^ { infty } a r ^ { n - 1 } = frac { a _ { } } { 1 - r ^ { } }$
$endgroup$
– Alex
Jan 28 at 19:35
add a comment |
$begingroup$
Thanks for the hint @Mindlack. I know very little about geometric series (and series in general) though. I guess this now should look like some expression for which the sum is well known. I will try to compute this using online solver (although it could not give me an answer for the original form).
$endgroup$
– Alex
Jan 28 at 19:27
$begingroup$
I got it now :), many thanks @Mindlack. Just needed to replace accordingly in $S _ { infty } = sum _ { n = 1 } ^ { infty } a r ^ { n - 1 } = frac { a _ { } } { 1 - r ^ { } }$
$endgroup$
– Alex
Jan 28 at 19:35
$begingroup$
Thanks for the hint @Mindlack. I know very little about geometric series (and series in general) though. I guess this now should look like some expression for which the sum is well known. I will try to compute this using online solver (although it could not give me an answer for the original form).
$endgroup$
– Alex
Jan 28 at 19:27
$begingroup$
Thanks for the hint @Mindlack. I know very little about geometric series (and series in general) though. I guess this now should look like some expression for which the sum is well known. I will try to compute this using online solver (although it could not give me an answer for the original form).
$endgroup$
– Alex
Jan 28 at 19:27
$begingroup$
I got it now :), many thanks @Mindlack. Just needed to replace accordingly in $S _ { infty } = sum _ { n = 1 } ^ { infty } a r ^ { n - 1 } = frac { a _ { } } { 1 - r ^ { } }$
$endgroup$
– Alex
Jan 28 at 19:35
$begingroup$
I got it now :), many thanks @Mindlack. Just needed to replace accordingly in $S _ { infty } = sum _ { n = 1 } ^ { infty } a r ^ { n - 1 } = frac { a _ { } } { 1 - r ^ { } }$
$endgroup$
– Alex
Jan 28 at 19:35
add a comment |
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