Proving this complicated limit! The 2019 Stack Overflow Developer Survey Results Are InWhat is the correct radius of convergence for $ln(1+x)$?Help in proving $lim limits_n to infty (a_n pm b_n) = a pm b$limit of a function involving $arcsinx$ and $ln(1+x)$Help with a limit problemLimit of a sum versus individualIf $lim_limitsntoinfty(a_nsum_i=1^na_i^2)=1$, then $lim_limitsntoinftysqrt[3](3n)a_n=1$Limit of infinite summationProving the limitLimit of a complex sumSwapping limit operators in a specific probability exampleTrouble with L'Hôpital Justification for Complicated Limit of Integral
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Proving this complicated limit!
The 2019 Stack Overflow Developer Survey Results Are InWhat is the correct radius of convergence for $ln(1+x)$?Help in proving $lim limits_n to infty (a_n pm b_n) = a pm b$limit of a function involving $arcsinx$ and $ln(1+x)$Help with a limit problemLimit of a sum versus individualIf $lim_limitsntoinfty(a_nsum_i=1^na_i^2)=1$, then $lim_limitsntoinftysqrt[3](3n)a_n=1$Limit of infinite summationProving the limitLimit of a complex sumSwapping limit operators in a specific probability exampleTrouble with L'Hôpital Justification for Complicated Limit of Integral
$begingroup$
I have to prove $limlimits_n to infty sum_limitsi=1^n frac(-1)^i+1i = ln2$.
The hints I have are:
$limlimits_n to infty sum_limitsi=1^n frac1i - ln(n) = c < infty$
$sum_limitsi=1^2n frac(-1)^i+1i = sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^2n frac12i $
I never calculated any limits with sum in the formula, nor I see how the tips are of any help.
Would appreciate a few hints and tips!
limits
edited Apr 6 at 15:40
EnlightenedFunky
84211022
asked Apr 6 at 15:33
TegernakoTegernako
948
$endgroup$
add a comment |
$begingroup$
I have to prove $limlimits_n to infty sum_limitsi=1^n frac(-1)^i+1i = ln2$.
The hints I have are:
$limlimits_n to infty sum_limitsi=1^n frac1i - ln(n) = c < infty$
$sum_limitsi=1^2n frac(-1)^i+1i = sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^2n frac12i $
I never calculated any limits with sum in the formula, nor I see how the tips are of any help.
Would appreciate a few hints and tips!
limits
edited Apr 6 at 15:40
EnlightenedFunky
84211022
asked Apr 6 at 15:33
TegernakoTegernako
948
$endgroup$
$begingroup$
See math.stackexchange.com/questions/1356517/…
$endgroup$
– lab bhattacharjee
Apr 6 at 15:37
$begingroup$
The second hint should be $sum_limitsi=1^2n frac(-1)^i+1i = sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^n frac12i$.
$endgroup$
– Robert Z
Apr 6 at 15:47
add a comment |
$begingroup$
I have to prove $limlimits_n to infty sum_limitsi=1^n frac(-1)^i+1i = ln2$.
The hints I have are:
$limlimits_n to infty sum_limitsi=1^n frac1i - ln(n) = c < infty$
$sum_limitsi=1^2n frac(-1)^i+1i = sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^2n frac12i $
I never calculated any limits with sum in the formula, nor I see how the tips are of any help.
Would appreciate a few hints and tips!
limits
edited Apr 6 at 15:40
EnlightenedFunky
84211022
asked Apr 6 at 15:33
TegernakoTegernako
948
$endgroup$
I have to prove $limlimits_n to infty sum_limitsi=1^n frac(-1)^i+1i = ln2$.
The hints I have are:
$limlimits_n to infty sum_limitsi=1^n frac1i - ln(n) = c < infty$
$sum_limitsi=1^2n frac(-1)^i+1i = sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^2n frac12i $
I never calculated any limits with sum in the formula, nor I see how the tips are of any help.
Would appreciate a few hints and tips!
limits
limits
edited Apr 6 at 15:40
EnlightenedFunky
84211022
asked Apr 6 at 15:33
TegernakoTegernako
948
edited Apr 6 at 15:40
EnlightenedFunky
84211022
asked Apr 6 at 15:33
TegernakoTegernako
948
edited Apr 6 at 15:40
EnlightenedFunky
84211022
edited Apr 6 at 15:40
EnlightenedFunky
84211022
edited Apr 6 at 15:40
EnlightenedFunky
84211022
84211022
asked Apr 6 at 15:33
TegernakoTegernako
948
asked Apr 6 at 15:33
TegernakoTegernako
948
asked Apr 6 at 15:33
TegernakoTegernako
948
948
$begingroup$
See math.stackexchange.com/questions/1356517/…
$endgroup$
– lab bhattacharjee
Apr 6 at 15:37
$begingroup$
The second hint should be $sum_limitsi=1^2n frac(-1)^i+1i = sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^n frac12i$.
$endgroup$
– Robert Z
Apr 6 at 15:47
add a comment |
$begingroup$
See math.stackexchange.com/questions/1356517/…
$endgroup$
– lab bhattacharjee
Apr 6 at 15:37
$begingroup$
The second hint should be $sum_limitsi=1^2n frac(-1)^i+1i = sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^n frac12i$.
$endgroup$
– Robert Z
Apr 6 at 15:47
$begingroup$
See math.stackexchange.com/questions/1356517/…
$endgroup$
– lab bhattacharjee
Apr 6 at 15:37
$begingroup$
See math.stackexchange.com/questions/1356517/…
$endgroup$
– lab bhattacharjee
Apr 6 at 15:37
$begingroup$
The second hint should be $sum_limitsi=1^2n frac(-1)^i+1i = sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^n frac12i$.
$endgroup$
– Robert Z
Apr 6 at 15:47
$begingroup$
The second hint should be $sum_limitsi=1^2n frac(-1)^i+1i = sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^n frac12i$.
$endgroup$
– Robert Z
Apr 6 at 15:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $a_n=sum_i=1^n frac1i - ln(n+1)$. Then the sequence $(a_n)_ngeq 1$ is increasing:
$$a_n+1-a_n=frac1n+1-lnleft(1+frac1n+1right)geq 0$$
and bounded above (show that part).
Then the sequence $(a_n)_n$ converges to a finite limit $c$.
Hence, by the second hint,
$$beginalign
sum_limitsi=1^2n frac(-1)^i+1i &= sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^n frac12i\
&=left(sum_limitsi=1^2n frac 1i-ln(2n+1)right) - left(sum_limitsi=1^n frac1i-ln(n+1)right)+lnleft(frac2n+1n+1right)\
&to c-c+ln(2)=ln(2).
endalign$$
Finally note that
$$sum_limitsi=1^2n+1 frac(-1)^i+1i
=sum_limitsi=1^2n frac(-1)^i+1i+
frac12n+1to ln(2)+0=ln(2).$$
answered Apr 6 at 15:42
Robert ZRobert Z
101k1072145
$endgroup$
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
$begingroup$
Let $a_n=sum_i=1^n frac1i - ln(n+1)$. Then the sequence $(a_n)_ngeq 1$ is increasing:
$$a_n+1-a_n=frac1n+1-lnleft(1+frac1n+1right)geq 0$$
and bounded above (show that part).
Then the sequence $(a_n)_n$ converges to a finite limit $c$.
Hence, by the second hint,
$$beginalign
sum_limitsi=1^2n frac(-1)^i+1i &= sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^n frac12i\
&=left(sum_limitsi=1^2n frac 1i-ln(2n+1)right) - left(sum_limitsi=1^n frac1i-ln(n+1)right)+lnleft(frac2n+1n+1right)\
&to c-c+ln(2)=ln(2).
endalign$$
Finally note that
$$sum_limitsi=1^2n+1 frac(-1)^i+1i
=sum_limitsi=1^2n frac(-1)^i+1i+
frac12n+1to ln(2)+0=ln(2).$$
answered Apr 6 at 15:42
Robert ZRobert Z
101k1072145
$endgroup$
add a comment |
$begingroup$
Let $a_n=sum_i=1^n frac1i - ln(n+1)$. Then the sequence $(a_n)_ngeq 1$ is increasing:
$$a_n+1-a_n=frac1n+1-lnleft(1+frac1n+1right)geq 0$$
and bounded above (show that part).
Then the sequence $(a_n)_n$ converges to a finite limit $c$.
Hence, by the second hint,
$$beginalign
sum_limitsi=1^2n frac(-1)^i+1i &= sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^n frac12i\
&=left(sum_limitsi=1^2n frac 1i-ln(2n+1)right) - left(sum_limitsi=1^n frac1i-ln(n+1)right)+lnleft(frac2n+1n+1right)\
&to c-c+ln(2)=ln(2).
endalign$$
Finally note that
$$sum_limitsi=1^2n+1 frac(-1)^i+1i
=sum_limitsi=1^2n frac(-1)^i+1i+
frac12n+1to ln(2)+0=ln(2).$$
answered Apr 6 at 15:42
Robert ZRobert Z
101k1072145
$endgroup$
add a comment |
$begingroup$
Let $a_n=sum_i=1^n frac1i - ln(n+1)$. Then the sequence $(a_n)_ngeq 1$ is increasing:
$$a_n+1-a_n=frac1n+1-lnleft(1+frac1n+1right)geq 0$$
and bounded above (show that part).
Then the sequence $(a_n)_n$ converges to a finite limit $c$.
Hence, by the second hint,
$$beginalign
sum_limitsi=1^2n frac(-1)^i+1i &= sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^n frac12i\
&=left(sum_limitsi=1^2n frac 1i-ln(2n+1)right) - left(sum_limitsi=1^n frac1i-ln(n+1)right)+lnleft(frac2n+1n+1right)\
&to c-c+ln(2)=ln(2).
endalign$$
Finally note that
$$sum_limitsi=1^2n+1 frac(-1)^i+1i
=sum_limitsi=1^2n frac(-1)^i+1i+
frac12n+1to ln(2)+0=ln(2).$$
answered Apr 6 at 15:42
Robert ZRobert Z
101k1072145
$endgroup$
Let $a_n=sum_i=1^n frac1i - ln(n+1)$. Then the sequence $(a_n)_ngeq 1$ is increasing:
$$a_n+1-a_n=frac1n+1-lnleft(1+frac1n+1right)geq 0$$
and bounded above (show that part).
Then the sequence $(a_n)_n$ converges to a finite limit $c$.
Hence, by the second hint,
$$beginalign
sum_limitsi=1^2n frac(-1)^i+1i &= sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^n frac12i\
&=left(sum_limitsi=1^2n frac 1i-ln(2n+1)right) - left(sum_limitsi=1^n frac1i-ln(n+1)right)+lnleft(frac2n+1n+1right)\
&to c-c+ln(2)=ln(2).
endalign$$
Finally note that
$$sum_limitsi=1^2n+1 frac(-1)^i+1i
=sum_limitsi=1^2n frac(-1)^i+1i+
frac12n+1to ln(2)+0=ln(2).$$
answered Apr 6 at 15:42
Robert ZRobert Z
101k1072145
edited Apr 6 at 16:56
answered Apr 6 at 15:42
Robert ZRobert Z
101k1072145
answered Apr 6 at 15:42
Robert ZRobert Z
101k1072145
answered Apr 6 at 15:42
Robert ZRobert Z
101k1072145
101k1072145
add a comment |
add a comment |
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$begingroup$
See math.stackexchange.com/questions/1356517/…
$endgroup$
– lab bhattacharjee
Apr 6 at 15:37
$begingroup$
The second hint should be $sum_limitsi=1^2n frac(-1)^i+1i = sum_limitsi=1^2n frac 1i - 2sum_limitsi=1^n frac12i$.
$endgroup$
– Robert Z
Apr 6 at 15:47