Evaluating series with factorial denominator (sanity check).
$begingroup$
Is my approach to evaluating this series correct?
$$sum_{n=1}^infty frac{n}{(n+1)!}$$
Has partial sum equivalent to:
$$S_m = sum_{n=1}^m frac{n}{(n+1)!} = sum_{j=2}^{m+1} frac{j-1}{j!} = sum_{j=2}^{m+1} frac{1}{(j-1)!} - sum_{j=2}^{m+1} frac{1}{j!} $$
For $j$ such that $m+1>j>2$ the terms of the left sum are cancelled by the terms of the right, leaving
$$ S_m =1-frac{1}{(m+1)!}$$
Hence $ lim_{mrightarrowinfty} S_m = 1$
Apologies for this one. The book I am using hasn't really offered anything on series with factorial denominators (yet). Thanks!
sequences-and-series
edited Jan 6 '13 at 0:19
Calvin Lin
36.3k349114
asked Jan 5 '13 at 23:54
conjecturesconjectures
534513
$endgroup$
add a comment |
$begingroup$
Is my approach to evaluating this series correct?
$$sum_{n=1}^infty frac{n}{(n+1)!}$$
Has partial sum equivalent to:
$$S_m = sum_{n=1}^m frac{n}{(n+1)!} = sum_{j=2}^{m+1} frac{j-1}{j!} = sum_{j=2}^{m+1} frac{1}{(j-1)!} - sum_{j=2}^{m+1} frac{1}{j!} $$
For $j$ such that $m+1>j>2$ the terms of the left sum are cancelled by the terms of the right, leaving
$$ S_m =1-frac{1}{(m+1)!}$$
Hence $ lim_{mrightarrowinfty} S_m = 1$
Apologies for this one. The book I am using hasn't really offered anything on series with factorial denominators (yet). Thanks!
sequences-and-series
edited Jan 6 '13 at 0:19
Calvin Lin
36.3k349114
asked Jan 5 '13 at 23:54
conjecturesconjectures
534513
$endgroup$
$begingroup$
Looks fine to me.
$endgroup$
– Brian M. Scott
Jan 5 '13 at 23:59
$begingroup$
I see nothing wrong. I'm editing to improve formatting, however.
$endgroup$
– Alex Becker
Jan 6 '13 at 0:03
$begingroup$
It should be $S_m$ everywhere
$endgroup$
– leo
Jan 6 '13 at 0:04
add a comment |
$begingroup$
Is my approach to evaluating this series correct?
$$sum_{n=1}^infty frac{n}{(n+1)!}$$
Has partial sum equivalent to:
$$S_m = sum_{n=1}^m frac{n}{(n+1)!} = sum_{j=2}^{m+1} frac{j-1}{j!} = sum_{j=2}^{m+1} frac{1}{(j-1)!} - sum_{j=2}^{m+1} frac{1}{j!} $$
For $j$ such that $m+1>j>2$ the terms of the left sum are cancelled by the terms of the right, leaving
$$ S_m =1-frac{1}{(m+1)!}$$
Hence $ lim_{mrightarrowinfty} S_m = 1$
Apologies for this one. The book I am using hasn't really offered anything on series with factorial denominators (yet). Thanks!
sequences-and-series
edited Jan 6 '13 at 0:19
Calvin Lin
36.3k349114
asked Jan 5 '13 at 23:54
conjecturesconjectures
534513
$endgroup$
Is my approach to evaluating this series correct?
$$sum_{n=1}^infty frac{n}{(n+1)!}$$
Has partial sum equivalent to:
$$S_m = sum_{n=1}^m frac{n}{(n+1)!} = sum_{j=2}^{m+1} frac{j-1}{j!} = sum_{j=2}^{m+1} frac{1}{(j-1)!} - sum_{j=2}^{m+1} frac{1}{j!} $$
For $j$ such that $m+1>j>2$ the terms of the left sum are cancelled by the terms of the right, leaving
$$ S_m =1-frac{1}{(m+1)!}$$
Hence $ lim_{mrightarrowinfty} S_m = 1$
Apologies for this one. The book I am using hasn't really offered anything on series with factorial denominators (yet). Thanks!
sequences-and-series
sequences-and-series
edited Jan 6 '13 at 0:19
Calvin Lin
36.3k349114
asked Jan 5 '13 at 23:54
conjecturesconjectures
534513
edited Jan 6 '13 at 0:19
Calvin Lin
36.3k349114
asked Jan 5 '13 at 23:54
conjecturesconjectures
534513
edited Jan 6 '13 at 0:19
Calvin Lin
36.3k349114
edited Jan 6 '13 at 0:19
Calvin Lin
36.3k349114
edited Jan 6 '13 at 0:19
Calvin Lin
36.3k349114
36.3k349114
asked Jan 5 '13 at 23:54
conjecturesconjectures
534513
asked Jan 5 '13 at 23:54
conjecturesconjectures
534513
asked Jan 5 '13 at 23:54
conjecturesconjectures
534513
534513
$begingroup$
Looks fine to me.
$endgroup$
– Brian M. Scott
Jan 5 '13 at 23:59
$begingroup$
I see nothing wrong. I'm editing to improve formatting, however.
$endgroup$
– Alex Becker
Jan 6 '13 at 0:03
$begingroup$
It should be $S_m$ everywhere
$endgroup$
– leo
Jan 6 '13 at 0:04
add a comment |
$begingroup$
Looks fine to me.
$endgroup$
– Brian M. Scott
Jan 5 '13 at 23:59
$begingroup$
I see nothing wrong. I'm editing to improve formatting, however.
$endgroup$
– Alex Becker
Jan 6 '13 at 0:03
$begingroup$
It should be $S_m$ everywhere
$endgroup$
– leo
Jan 6 '13 at 0:04
$begingroup$
Looks fine to me.
$endgroup$
– Brian M. Scott
Jan 5 '13 at 23:59
$begingroup$
Looks fine to me.
$endgroup$
– Brian M. Scott
Jan 5 '13 at 23:59
$begingroup$
I see nothing wrong. I'm editing to improve formatting, however.
$endgroup$
– Alex Becker
Jan 6 '13 at 0:03
$begingroup$
I see nothing wrong. I'm editing to improve formatting, however.
$endgroup$
– Alex Becker
Jan 6 '13 at 0:03
$begingroup$
It should be $S_m$ everywhere
$endgroup$
– leo
Jan 6 '13 at 0:04
$begingroup$
It should be $S_m$ everywhere
$endgroup$
– leo
Jan 6 '13 at 0:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your solution is not only correct but also a very nice one.
Once you know the closed form for the partial sum, you can also prove it by induction, starting with $S_1=1-1/2=1/2$ and taking the induction step
$$
begin{align}
S_{m+1}&=S_m+frac{m+1}{(m+2)!}
\
&=1-frac1{(m+1)!}+frac{m+1}{(m+2)!}
\
&=1+frac{m+1-(m+2)}{(m+2)!}
\
&=1-frac1{(m+2)!};.
end{align}
$$
answered Jan 6 '13 at 8:02
jorikijoriki
170k10183343
$endgroup$
$begingroup$
Cheers Joriki - hadn't actually thought of trying it by induction.
$endgroup$
– conjectures
Jan 6 '13 at 10:14
add a comment |
$begingroup$
$$sum_{n=1}^infty frac{n}{(n+1)!}$$= $$sum_{n=1}^infty frac{n+1-1}{(n+1)!}$$=
$$sum_{n=1}^infty frac{1}{n!}-frac{1}{(n+1)!}$$
answered Dec 19 '18 at 15:15
Kostas GiatzoKostas Giatzo
12
$endgroup$
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
$begingroup$
Your solution is not only correct but also a very nice one.
Once you know the closed form for the partial sum, you can also prove it by induction, starting with $S_1=1-1/2=1/2$ and taking the induction step
$$
begin{align}
S_{m+1}&=S_m+frac{m+1}{(m+2)!}
\
&=1-frac1{(m+1)!}+frac{m+1}{(m+2)!}
\
&=1+frac{m+1-(m+2)}{(m+2)!}
\
&=1-frac1{(m+2)!};.
end{align}
$$
answered Jan 6 '13 at 8:02
jorikijoriki
170k10183343
$endgroup$
$begingroup$
Cheers Joriki - hadn't actually thought of trying it by induction.
$endgroup$
– conjectures
Jan 6 '13 at 10:14
add a comment |
$begingroup$
Your solution is not only correct but also a very nice one.
Once you know the closed form for the partial sum, you can also prove it by induction, starting with $S_1=1-1/2=1/2$ and taking the induction step
$$
begin{align}
S_{m+1}&=S_m+frac{m+1}{(m+2)!}
\
&=1-frac1{(m+1)!}+frac{m+1}{(m+2)!}
\
&=1+frac{m+1-(m+2)}{(m+2)!}
\
&=1-frac1{(m+2)!};.
end{align}
$$
answered Jan 6 '13 at 8:02
jorikijoriki
170k10183343
$endgroup$
$begingroup$
Cheers Joriki - hadn't actually thought of trying it by induction.
$endgroup$
– conjectures
Jan 6 '13 at 10:14
add a comment |
$begingroup$
Your solution is not only correct but also a very nice one.
Once you know the closed form for the partial sum, you can also prove it by induction, starting with $S_1=1-1/2=1/2$ and taking the induction step
$$
begin{align}
S_{m+1}&=S_m+frac{m+1}{(m+2)!}
\
&=1-frac1{(m+1)!}+frac{m+1}{(m+2)!}
\
&=1+frac{m+1-(m+2)}{(m+2)!}
\
&=1-frac1{(m+2)!};.
end{align}
$$
answered Jan 6 '13 at 8:02
jorikijoriki
170k10183343
$endgroup$
Your solution is not only correct but also a very nice one.
Once you know the closed form for the partial sum, you can also prove it by induction, starting with $S_1=1-1/2=1/2$ and taking the induction step
$$
begin{align}
S_{m+1}&=S_m+frac{m+1}{(m+2)!}
\
&=1-frac1{(m+1)!}+frac{m+1}{(m+2)!}
\
&=1+frac{m+1-(m+2)}{(m+2)!}
\
&=1-frac1{(m+2)!};.
end{align}
$$
answered Jan 6 '13 at 8:02
jorikijoriki
170k10183343
answered Jan 6 '13 at 8:02
jorikijoriki
170k10183343
answered Jan 6 '13 at 8:02
jorikijoriki
170k10183343
answered Jan 6 '13 at 8:02
jorikijoriki
170k10183343
170k10183343
$begingroup$
Cheers Joriki - hadn't actually thought of trying it by induction.
$endgroup$
– conjectures
Jan 6 '13 at 10:14
add a comment |
$begingroup$
Cheers Joriki - hadn't actually thought of trying it by induction.
$endgroup$
– conjectures
Jan 6 '13 at 10:14
$begingroup$
Cheers Joriki - hadn't actually thought of trying it by induction.
$endgroup$
– conjectures
Jan 6 '13 at 10:14
$begingroup$
Cheers Joriki - hadn't actually thought of trying it by induction.
$endgroup$
– conjectures
Jan 6 '13 at 10:14
add a comment |
$begingroup$
$$sum_{n=1}^infty frac{n}{(n+1)!}$$= $$sum_{n=1}^infty frac{n+1-1}{(n+1)!}$$=
$$sum_{n=1}^infty frac{1}{n!}-frac{1}{(n+1)!}$$
answered Dec 19 '18 at 15:15
Kostas GiatzoKostas Giatzo
12
$endgroup$
add a comment |
$begingroup$
$$sum_{n=1}^infty frac{n}{(n+1)!}$$= $$sum_{n=1}^infty frac{n+1-1}{(n+1)!}$$=
$$sum_{n=1}^infty frac{1}{n!}-frac{1}{(n+1)!}$$
answered Dec 19 '18 at 15:15
Kostas GiatzoKostas Giatzo
12
$endgroup$
add a comment |
$begingroup$
$$sum_{n=1}^infty frac{n}{(n+1)!}$$= $$sum_{n=1}^infty frac{n+1-1}{(n+1)!}$$=
$$sum_{n=1}^infty frac{1}{n!}-frac{1}{(n+1)!}$$
answered Dec 19 '18 at 15:15
Kostas GiatzoKostas Giatzo
12
$endgroup$
$$sum_{n=1}^infty frac{n}{(n+1)!}$$= $$sum_{n=1}^infty frac{n+1-1}{(n+1)!}$$=
$$sum_{n=1}^infty frac{1}{n!}-frac{1}{(n+1)!}$$
answered Dec 19 '18 at 15:15
Kostas GiatzoKostas Giatzo
12
answered Dec 19 '18 at 15:15
Kostas GiatzoKostas Giatzo
12
answered Dec 19 '18 at 15:15
Kostas GiatzoKostas Giatzo
12
answered Dec 19 '18 at 15:15
Kostas GiatzoKostas Giatzo
12
12
add a comment |
add a comment |
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$begingroup$
Looks fine to me.
$endgroup$
– Brian M. Scott
Jan 5 '13 at 23:59
$begingroup$
I see nothing wrong. I'm editing to improve formatting, however.
$endgroup$
– Alex Becker
Jan 6 '13 at 0:03
$begingroup$
It should be $S_m$ everywhere
$endgroup$
– leo
Jan 6 '13 at 0:04