Finding limit of a (Laurent?) series
Clash Royale CLAN TAG#URR8PPP
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I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
$1 + frac11 + 2 + frac11 + 2 + 3 + ... + frac11 + 2 + 3 + ... + n$
The task is to find the limit.
sequences-and-series limits
edited Feb 17 at 15:29
Max
631317
asked Feb 17 at 14:25
RoseRose
216
$endgroup$
add a comment |
$begingroup$
I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
$1 + frac11 + 2 + frac11 + 2 + 3 + ... + frac11 + 2 + 3 + ... + n$
The task is to find the limit.
sequences-and-series limits
edited Feb 17 at 15:29
Max
631317
asked Feb 17 at 14:25
RoseRose
216
$endgroup$
add a comment |
$begingroup$
I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
$1 + frac11 + 2 + frac11 + 2 + 3 + ... + frac11 + 2 + 3 + ... + n$
The task is to find the limit.
sequences-and-series limits
edited Feb 17 at 15:29
Max
631317
asked Feb 17 at 14:25
RoseRose
216
$endgroup$
I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
$1 + frac11 + 2 + frac11 + 2 + 3 + ... + frac11 + 2 + 3 + ... + n$
The task is to find the limit.
sequences-and-series limits
sequences-and-series limits
edited Feb 17 at 15:29
Max
631317
asked Feb 17 at 14:25
RoseRose
216
edited Feb 17 at 15:29
Max
631317
asked Feb 17 at 14:25
RoseRose
216
edited Feb 17 at 15:29
Max
631317
edited Feb 17 at 15:29
Max
631317
edited Feb 17 at 15:29
Max
631317
631317
asked Feb 17 at 14:25
RoseRose
216
asked Feb 17 at 14:25
RoseRose
216
asked Feb 17 at 14:25
RoseRose
216
216
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
We have $$1+2+...+k=frack(k+1)2$$
So the sum you need to compute is
$$beginsplit
sum_k=1^n frac 2k(k+1) &= sum_k=1^n 2left ( frac 1 k - frac 1 k+1 right )\
&=2-frac 2 n+1\
&=frac 2n n+1
endsplit$$
Now you can take the limit.
edited Feb 17 at 19:03
RQM
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,92519
$endgroup$
add a comment |
$begingroup$
Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
$$1+2+cdots+n=fracn(n+1)2$$
Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
$$
frac2n(n+1)=2(frac1n-frac1n+1)$$
After that, take the limit.
answered Feb 17 at 14:34
weilam06weilam06
30412
$endgroup$
add a comment |
$begingroup$
Hint: Use the fact that $$1+2+...+n=fracn(n+1)2$$
The series then becomes $$2sumlimits_n=1^infty left(frac1n - frac1n+1 right)$$ which is a telescoping series.
answered Feb 17 at 14:33
Haris GusicHaris Gusic
2,710423
$endgroup$
add a comment |
$begingroup$
As $1+2+ldots +n = fracn2(n+1)$ you are looking for the sum of the series
$S = sum_n=1^infty frac1frack2(k+1) = 2 sum_k=1^infty frac1k(k+1)$.
Nos separating you get that $S_n = 2sum_k=1^n frac1k - frac1k+1 = 2(1 - frac1n+1)$. Now taking limit when $n rightarrow infty$ you get that $S=2$
answered Feb 17 at 14:36
JoseSquareJoseSquare
867115
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add a comment |
$begingroup$
$smalla_2=(1+2)^-1, a_3= (1+2+3)^-1,....$
$smalla_k=(1+2+...k)^-1 = 2(k(k+1))^-1=2(1/k -1/(k+1))$
Telescopic sum
$1+sum_k=2^infty a_k =?$
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have $$1+2+...+k=frack(k+1)2$$
So the sum you need to compute is
$$beginsplit
sum_k=1^n frac 2k(k+1) &= sum_k=1^n 2left ( frac 1 k - frac 1 k+1 right )\
&=2-frac 2 n+1\
&=frac 2n n+1
endsplit$$
Now you can take the limit.
edited Feb 17 at 19:03
RQM
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,92519
$endgroup$
add a comment |
$begingroup$
We have $$1+2+...+k=frack(k+1)2$$
So the sum you need to compute is
$$beginsplit
sum_k=1^n frac 2k(k+1) &= sum_k=1^n 2left ( frac 1 k - frac 1 k+1 right )\
&=2-frac 2 n+1\
&=frac 2n n+1
endsplit$$
Now you can take the limit.
edited Feb 17 at 19:03
RQM
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,92519
$endgroup$
add a comment |
$begingroup$
We have $$1+2+...+k=frack(k+1)2$$
So the sum you need to compute is
$$beginsplit
sum_k=1^n frac 2k(k+1) &= sum_k=1^n 2left ( frac 1 k - frac 1 k+1 right )\
&=2-frac 2 n+1\
&=frac 2n n+1
endsplit$$
Now you can take the limit.
edited Feb 17 at 19:03
RQM
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,92519
$endgroup$
We have $$1+2+...+k=frack(k+1)2$$
So the sum you need to compute is
$$beginsplit
sum_k=1^n frac 2k(k+1) &= sum_k=1^n 2left ( frac 1 k - frac 1 k+1 right )\
&=2-frac 2 n+1\
&=frac 2n n+1
endsplit$$
Now you can take the limit.
edited Feb 17 at 19:03
RQM
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,92519
edited Feb 17 at 19:03
RQM
1213
edited Feb 17 at 19:03
RQM
1213
edited Feb 17 at 19:03
RQM
1213
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,92519
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,92519
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,92519
2,92519
add a comment |
add a comment |
$begingroup$
Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
$$1+2+cdots+n=fracn(n+1)2$$
Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
$$
frac2n(n+1)=2(frac1n-frac1n+1)$$
After that, take the limit.
answered Feb 17 at 14:34
weilam06weilam06
30412
$endgroup$
add a comment |
$begingroup$
Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
$$1+2+cdots+n=fracn(n+1)2$$
Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
$$
frac2n(n+1)=2(frac1n-frac1n+1)$$
After that, take the limit.
answered Feb 17 at 14:34
weilam06weilam06
30412
$endgroup$
add a comment |
$begingroup$
Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
$$1+2+cdots+n=fracn(n+1)2$$
Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
$$
frac2n(n+1)=2(frac1n-frac1n+1)$$
After that, take the limit.
answered Feb 17 at 14:34
weilam06weilam06
30412
$endgroup$
Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
$$1+2+cdots+n=fracn(n+1)2$$
Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
$$
frac2n(n+1)=2(frac1n-frac1n+1)$$
After that, take the limit.
answered Feb 17 at 14:34
weilam06weilam06
30412
answered Feb 17 at 14:34
weilam06weilam06
30412
answered Feb 17 at 14:34
weilam06weilam06
30412
answered Feb 17 at 14:34
weilam06weilam06
30412
30412
add a comment |
add a comment |
$begingroup$
Hint: Use the fact that $$1+2+...+n=fracn(n+1)2$$
The series then becomes $$2sumlimits_n=1^infty left(frac1n - frac1n+1 right)$$ which is a telescoping series.
answered Feb 17 at 14:33
Haris GusicHaris Gusic
2,710423
$endgroup$
add a comment |
$begingroup$
Hint: Use the fact that $$1+2+...+n=fracn(n+1)2$$
The series then becomes $$2sumlimits_n=1^infty left(frac1n - frac1n+1 right)$$ which is a telescoping series.
answered Feb 17 at 14:33
Haris GusicHaris Gusic
2,710423
$endgroup$
add a comment |
$begingroup$
Hint: Use the fact that $$1+2+...+n=fracn(n+1)2$$
The series then becomes $$2sumlimits_n=1^infty left(frac1n - frac1n+1 right)$$ which is a telescoping series.
answered Feb 17 at 14:33
Haris GusicHaris Gusic
2,710423
$endgroup$
Hint: Use the fact that $$1+2+...+n=fracn(n+1)2$$
The series then becomes $$2sumlimits_n=1^infty left(frac1n - frac1n+1 right)$$ which is a telescoping series.
answered Feb 17 at 14:33
Haris GusicHaris Gusic
2,710423
answered Feb 17 at 14:33
Haris GusicHaris Gusic
2,710423
answered Feb 17 at 14:33
Haris GusicHaris Gusic
2,710423
answered Feb 17 at 14:33
Haris GusicHaris Gusic
2,710423
2,710423
add a comment |
add a comment |
$begingroup$
As $1+2+ldots +n = fracn2(n+1)$ you are looking for the sum of the series
$S = sum_n=1^infty frac1frack2(k+1) = 2 sum_k=1^infty frac1k(k+1)$.
Nos separating you get that $S_n = 2sum_k=1^n frac1k - frac1k+1 = 2(1 - frac1n+1)$. Now taking limit when $n rightarrow infty$ you get that $S=2$
answered Feb 17 at 14:36
JoseSquareJoseSquare
867115
$endgroup$
add a comment |
$begingroup$
As $1+2+ldots +n = fracn2(n+1)$ you are looking for the sum of the series
$S = sum_n=1^infty frac1frack2(k+1) = 2 sum_k=1^infty frac1k(k+1)$.
Nos separating you get that $S_n = 2sum_k=1^n frac1k - frac1k+1 = 2(1 - frac1n+1)$. Now taking limit when $n rightarrow infty$ you get that $S=2$
answered Feb 17 at 14:36
JoseSquareJoseSquare
867115
$endgroup$
add a comment |
$begingroup$
As $1+2+ldots +n = fracn2(n+1)$ you are looking for the sum of the series
$S = sum_n=1^infty frac1frack2(k+1) = 2 sum_k=1^infty frac1k(k+1)$.
Nos separating you get that $S_n = 2sum_k=1^n frac1k - frac1k+1 = 2(1 - frac1n+1)$. Now taking limit when $n rightarrow infty$ you get that $S=2$
answered Feb 17 at 14:36
JoseSquareJoseSquare
867115
$endgroup$
As $1+2+ldots +n = fracn2(n+1)$ you are looking for the sum of the series
$S = sum_n=1^infty frac1frack2(k+1) = 2 sum_k=1^infty frac1k(k+1)$.
Nos separating you get that $S_n = 2sum_k=1^n frac1k - frac1k+1 = 2(1 - frac1n+1)$. Now taking limit when $n rightarrow infty$ you get that $S=2$
answered Feb 17 at 14:36
JoseSquareJoseSquare
867115
answered Feb 17 at 14:36
JoseSquareJoseSquare
867115
answered Feb 17 at 14:36
JoseSquareJoseSquare
867115
answered Feb 17 at 14:36
JoseSquareJoseSquare
867115
867115
add a comment |
add a comment |
$begingroup$
$smalla_2=(1+2)^-1, a_3= (1+2+3)^-1,....$
$smalla_k=(1+2+...k)^-1 = 2(k(k+1))^-1=2(1/k -1/(k+1))$
Telescopic sum
$1+sum_k=2^infty a_k =?$
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
$begingroup$
$smalla_2=(1+2)^-1, a_3= (1+2+3)^-1,....$
$smalla_k=(1+2+...k)^-1 = 2(k(k+1))^-1=2(1/k -1/(k+1))$
Telescopic sum
$1+sum_k=2^infty a_k =?$
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
$begingroup$
$smalla_2=(1+2)^-1, a_3= (1+2+3)^-1,....$
$smalla_k=(1+2+...k)^-1 = 2(k(k+1))^-1=2(1/k -1/(k+1))$
Telescopic sum
$1+sum_k=2^infty a_k =?$
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
$smalla_2=(1+2)^-1, a_3= (1+2+3)^-1,....$
$smalla_k=(1+2+...k)^-1 = 2(k(k+1))^-1=2(1/k -1/(k+1))$
Telescopic sum
$1+sum_k=2^infty a_k =?$
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.5k2822
edited Feb 17 at 16:50
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.5k2822
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.5k2822
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.5k2822
11.5k2822
add a comment |
add a comment |
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