Is the sum of a convergent series from infinity to infinity always zero?
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Suppose that $Sigma_n=1^infty a_n$ converges. Then $Sigma_n=k^infty a_n$ also converges for any $k>1$. So my question is, in such a case, does the limit of $Sigma_n=k^infty a_n$ as $k$ goes to $infty$ always equal $0$?
I think the answer is yes, because $Sigma_n=k^infty a_n=Sigma_n=1^infty a_n-Sigma_n=1^k-1 a_n$, and the right hand side goes to $Sigma_n=1^infty a_n-Sigma_n=1^infty a_n=0$ as $k$ goes to $infty$. But I just wanted to double check.
calculus sequences-and-series proof-verification convergence
edited Aug 31 at 1:11
gt6989b
30.7k22248
asked Aug 31 at 1:01
Keshav Srinivasan
1,85011339
add a comment |Â
up vote
4
down vote
favorite
Suppose that $Sigma_n=1^infty a_n$ converges. Then $Sigma_n=k^infty a_n$ also converges for any $k>1$. So my question is, in such a case, does the limit of $Sigma_n=k^infty a_n$ as $k$ goes to $infty$ always equal $0$?
I think the answer is yes, because $Sigma_n=k^infty a_n=Sigma_n=1^infty a_n-Sigma_n=1^k-1 a_n$, and the right hand side goes to $Sigma_n=1^infty a_n-Sigma_n=1^infty a_n=0$ as $k$ goes to $infty$. But I just wanted to double check.
calculus sequences-and-series proof-verification convergence
edited Aug 31 at 1:11
gt6989b
30.7k22248
asked Aug 31 at 1:01
Keshav Srinivasan
1,85011339
2
you are right. :)
– Ixion
Aug 31 at 1:04
2
That is correct.
– Jair Taylor
Aug 31 at 1:05
And, in fact, this is often a useful property for convergent series.
– GEdgar
Aug 31 at 9:53
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Suppose that $Sigma_n=1^infty a_n$ converges. Then $Sigma_n=k^infty a_n$ also converges for any $k>1$. So my question is, in such a case, does the limit of $Sigma_n=k^infty a_n$ as $k$ goes to $infty$ always equal $0$?
I think the answer is yes, because $Sigma_n=k^infty a_n=Sigma_n=1^infty a_n-Sigma_n=1^k-1 a_n$, and the right hand side goes to $Sigma_n=1^infty a_n-Sigma_n=1^infty a_n=0$ as $k$ goes to $infty$. But I just wanted to double check.
calculus sequences-and-series proof-verification convergence
edited Aug 31 at 1:11
gt6989b
30.7k22248
asked Aug 31 at 1:01
Keshav Srinivasan
1,85011339
Suppose that $Sigma_n=1^infty a_n$ converges. Then $Sigma_n=k^infty a_n$ also converges for any $k>1$. So my question is, in such a case, does the limit of $Sigma_n=k^infty a_n$ as $k$ goes to $infty$ always equal $0$?
I think the answer is yes, because $Sigma_n=k^infty a_n=Sigma_n=1^infty a_n-Sigma_n=1^k-1 a_n$, and the right hand side goes to $Sigma_n=1^infty a_n-Sigma_n=1^infty a_n=0$ as $k$ goes to $infty$. But I just wanted to double check.
calculus sequences-and-series proof-verification convergence
calculus sequences-and-series proof-verification convergence
edited Aug 31 at 1:11
gt6989b
30.7k22248
asked Aug 31 at 1:01
Keshav Srinivasan
1,85011339
edited Aug 31 at 1:11
gt6989b
30.7k22248
asked Aug 31 at 1:01
Keshav Srinivasan
1,85011339
edited Aug 31 at 1:11
gt6989b
30.7k22248
edited Aug 31 at 1:11
gt6989b
30.7k22248
edited Aug 31 at 1:11
gt6989b
30.7k22248
30.7k22248
asked Aug 31 at 1:01
Keshav Srinivasan
1,85011339
asked Aug 31 at 1:01
Keshav Srinivasan
1,85011339
asked Aug 31 at 1:01
Keshav Srinivasan
1,85011339
1,85011339
2
you are right. :)
– Ixion
Aug 31 at 1:04
2
That is correct.
– Jair Taylor
Aug 31 at 1:05
And, in fact, this is often a useful property for convergent series.
– GEdgar
Aug 31 at 9:53
add a comment |Â
2
you are right. :)
– Ixion
Aug 31 at 1:04
2
That is correct.
– Jair Taylor
Aug 31 at 1:05
And, in fact, this is often a useful property for convergent series.
– GEdgar
Aug 31 at 9:53
2
2
you are right. :)
– Ixion
Aug 31 at 1:04
you are right. :)
– Ixion
Aug 31 at 1:04
2
2
That is correct.
– Jair Taylor
Aug 31 at 1:05
That is correct.
– Jair Taylor
Aug 31 at 1:05
And, in fact, this is often a useful property for convergent series.
– GEdgar
Aug 31 at 9:53
And, in fact, this is often a useful property for convergent series.
– GEdgar
Aug 31 at 9:53
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Yes it does.
Note that if $$ Sigma_n=1^infty =S$$ then $$Sigma_n=k^infty a_n=Sigma_n=1^infty a_n-Sigma_n=1^k-1 a_n =S-Sigma_n=1^k-1 a_n$$
Thus if $kto $$infty$ we get $$ lim_kto inftySigma_n=k^infty a_n =S-S=0$$
answered Aug 31 at 1:33
Mohammad Riazi-Kermani
31.1k41853
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes it does.
Note that if $$ Sigma_n=1^infty =S$$ then $$Sigma_n=k^infty a_n=Sigma_n=1^infty a_n-Sigma_n=1^k-1 a_n =S-Sigma_n=1^k-1 a_n$$
Thus if $kto $$infty$ we get $$ lim_kto inftySigma_n=k^infty a_n =S-S=0$$
answered Aug 31 at 1:33
Mohammad Riazi-Kermani
31.1k41853
add a comment |Â
up vote
1
down vote
accepted
Yes it does.
Note that if $$ Sigma_n=1^infty =S$$ then $$Sigma_n=k^infty a_n=Sigma_n=1^infty a_n-Sigma_n=1^k-1 a_n =S-Sigma_n=1^k-1 a_n$$
Thus if $kto $$infty$ we get $$ lim_kto inftySigma_n=k^infty a_n =S-S=0$$
answered Aug 31 at 1:33
Mohammad Riazi-Kermani
31.1k41853
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes it does.
Note that if $$ Sigma_n=1^infty =S$$ then $$Sigma_n=k^infty a_n=Sigma_n=1^infty a_n-Sigma_n=1^k-1 a_n =S-Sigma_n=1^k-1 a_n$$
Thus if $kto $$infty$ we get $$ lim_kto inftySigma_n=k^infty a_n =S-S=0$$
answered Aug 31 at 1:33
Mohammad Riazi-Kermani
31.1k41853
Yes it does.
Note that if $$ Sigma_n=1^infty =S$$ then $$Sigma_n=k^infty a_n=Sigma_n=1^infty a_n-Sigma_n=1^k-1 a_n =S-Sigma_n=1^k-1 a_n$$
Thus if $kto $$infty$ we get $$ lim_kto inftySigma_n=k^infty a_n =S-S=0$$
answered Aug 31 at 1:33
Mohammad Riazi-Kermani
31.1k41853
answered Aug 31 at 1:33
Mohammad Riazi-Kermani
31.1k41853
answered Aug 31 at 1:33
Mohammad Riazi-Kermani
31.1k41853
answered Aug 31 at 1:33
Mohammad Riazi-Kermani
31.1k41853
31.1k41853
add a comment |Â
add a comment |Â
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2
you are right. :)
– Ixion
Aug 31 at 1:04
2
That is correct.
– Jair Taylor
Aug 31 at 1:05
And, in fact, this is often a useful property for convergent series.
– GEdgar
Aug 31 at 9:53