Does $sum_{n=1}^{infty} frac{1}{sqrt{n}-frac{2}3}$ converge or diverge?
$begingroup$
Does this series converge or diverge?
$$sum_{n=1}^{infty} frac{1}{sqrt{n}-frac{2}3}$$
I tried using the limit comparison test with $frac{1}{sqrt{n}}$, which diverges.
$$lim_{ntoinfty}{frac{{sqrt{n}}}{sqrt{n}-frac{2}3}}=1$$
Then the series diverges, is this right or I'm wrong?
sequences-and-series proof-verification convergence
edited Dec 22 '18 at 8:38
choco_addicted
8,06261847
asked Dec 5 '18 at 20:18
iggykimiiggykimi
1179
$endgroup$
add a comment |
$begingroup$
Does this series converge or diverge?
$$sum_{n=1}^{infty} frac{1}{sqrt{n}-frac{2}3}$$
I tried using the limit comparison test with $frac{1}{sqrt{n}}$, which diverges.
$$lim_{ntoinfty}{frac{{sqrt{n}}}{sqrt{n}-frac{2}3}}=1$$
Then the series diverges, is this right or I'm wrong?
sequences-and-series proof-verification convergence
edited Dec 22 '18 at 8:38
choco_addicted
8,06261847
asked Dec 5 '18 at 20:18
iggykimiiggykimi
1179
$endgroup$
$begingroup$
looks good to me
$endgroup$
– gt6989b
Dec 5 '18 at 20:19
add a comment |
$begingroup$
Does this series converge or diverge?
$$sum_{n=1}^{infty} frac{1}{sqrt{n}-frac{2}3}$$
I tried using the limit comparison test with $frac{1}{sqrt{n}}$, which diverges.
$$lim_{ntoinfty}{frac{{sqrt{n}}}{sqrt{n}-frac{2}3}}=1$$
Then the series diverges, is this right or I'm wrong?
sequences-and-series proof-verification convergence
edited Dec 22 '18 at 8:38
choco_addicted
8,06261847
asked Dec 5 '18 at 20:18
iggykimiiggykimi
1179
$endgroup$
Does this series converge or diverge?
$$sum_{n=1}^{infty} frac{1}{sqrt{n}-frac{2}3}$$
I tried using the limit comparison test with $frac{1}{sqrt{n}}$, which diverges.
$$lim_{ntoinfty}{frac{{sqrt{n}}}{sqrt{n}-frac{2}3}}=1$$
Then the series diverges, is this right or I'm wrong?
sequences-and-series proof-verification convergence
sequences-and-series proof-verification convergence
edited Dec 22 '18 at 8:38
choco_addicted
8,06261847
asked Dec 5 '18 at 20:18
iggykimiiggykimi
1179
edited Dec 22 '18 at 8:38
choco_addicted
8,06261847
asked Dec 5 '18 at 20:18
iggykimiiggykimi
1179
edited Dec 22 '18 at 8:38
choco_addicted
8,06261847
edited Dec 22 '18 at 8:38
choco_addicted
8,06261847
edited Dec 22 '18 at 8:38
choco_addicted
8,06261847
8,06261847
asked Dec 5 '18 at 20:18
iggykimiiggykimi
1179
asked Dec 5 '18 at 20:18
iggykimiiggykimi
1179
asked Dec 5 '18 at 20:18
iggykimiiggykimi
1179
1179
$begingroup$
looks good to me
$endgroup$
– gt6989b
Dec 5 '18 at 20:19
add a comment |
$begingroup$
looks good to me
$endgroup$
– gt6989b
Dec 5 '18 at 20:19
$begingroup$
looks good to me
$endgroup$
– gt6989b
Dec 5 '18 at 20:19
$begingroup$
looks good to me
$endgroup$
– gt6989b
Dec 5 '18 at 20:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes it is absolutely right, indeed note that
$$dfrac{1}{sqrt{n}-frac{2}3}sim dfrac{1}{sqrt{n}}$$
and the latter diverges for p test.
As an alternative by direct comparison test
$$dfrac{1}{sqrt{n}-frac{2}3}ge dfrac{1}{sqrt{n}}$$
answered Dec 5 '18 at 20:19
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
Yeah, just as another answer shows, you can see the convergence of such a sum of sequence does not matter w.r.t first several terms.
Then, you can see
$frac{1}{sqrt{n} -frac{2}{3}} sim frac{1}{sqrt{n}} = frac{1}{n^{1/2}}$
notice that for those power less than or equals 1,(here it's 1/2), it's a diverge sequence.
(You may refer to any analysis book for this result.)
answered Dec 5 '18 at 20:24
汪铈达汪铈达
111
$endgroup$
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
$begingroup$
Yes it is absolutely right, indeed note that
$$dfrac{1}{sqrt{n}-frac{2}3}sim dfrac{1}{sqrt{n}}$$
and the latter diverges for p test.
As an alternative by direct comparison test
$$dfrac{1}{sqrt{n}-frac{2}3}ge dfrac{1}{sqrt{n}}$$
answered Dec 5 '18 at 20:19
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
Yes it is absolutely right, indeed note that
$$dfrac{1}{sqrt{n}-frac{2}3}sim dfrac{1}{sqrt{n}}$$
and the latter diverges for p test.
As an alternative by direct comparison test
$$dfrac{1}{sqrt{n}-frac{2}3}ge dfrac{1}{sqrt{n}}$$
answered Dec 5 '18 at 20:19
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
Yes it is absolutely right, indeed note that
$$dfrac{1}{sqrt{n}-frac{2}3}sim dfrac{1}{sqrt{n}}$$
and the latter diverges for p test.
As an alternative by direct comparison test
$$dfrac{1}{sqrt{n}-frac{2}3}ge dfrac{1}{sqrt{n}}$$
answered Dec 5 '18 at 20:19
gimusigimusi
1
$endgroup$
Yes it is absolutely right, indeed note that
$$dfrac{1}{sqrt{n}-frac{2}3}sim dfrac{1}{sqrt{n}}$$
and the latter diverges for p test.
As an alternative by direct comparison test
$$dfrac{1}{sqrt{n}-frac{2}3}ge dfrac{1}{sqrt{n}}$$
answered Dec 5 '18 at 20:19
gimusigimusi
1
answered Dec 5 '18 at 20:19
gimusigimusi
1
answered Dec 5 '18 at 20:19
gimusigimusi
1
answered Dec 5 '18 at 20:19
gimusigimusi
1
1
add a comment |
add a comment |
$begingroup$
Yeah, just as another answer shows, you can see the convergence of such a sum of sequence does not matter w.r.t first several terms.
Then, you can see
$frac{1}{sqrt{n} -frac{2}{3}} sim frac{1}{sqrt{n}} = frac{1}{n^{1/2}}$
notice that for those power less than or equals 1,(here it's 1/2), it's a diverge sequence.
(You may refer to any analysis book for this result.)
answered Dec 5 '18 at 20:24
汪铈达汪铈达
111
$endgroup$
add a comment |
$begingroup$
Yeah, just as another answer shows, you can see the convergence of such a sum of sequence does not matter w.r.t first several terms.
Then, you can see
$frac{1}{sqrt{n} -frac{2}{3}} sim frac{1}{sqrt{n}} = frac{1}{n^{1/2}}$
notice that for those power less than or equals 1,(here it's 1/2), it's a diverge sequence.
(You may refer to any analysis book for this result.)
answered Dec 5 '18 at 20:24
汪铈达汪铈达
111
$endgroup$
add a comment |
$begingroup$
Yeah, just as another answer shows, you can see the convergence of such a sum of sequence does not matter w.r.t first several terms.
Then, you can see
$frac{1}{sqrt{n} -frac{2}{3}} sim frac{1}{sqrt{n}} = frac{1}{n^{1/2}}$
notice that for those power less than or equals 1,(here it's 1/2), it's a diverge sequence.
(You may refer to any analysis book for this result.)
answered Dec 5 '18 at 20:24
汪铈达汪铈达
111
$endgroup$
Yeah, just as another answer shows, you can see the convergence of such a sum of sequence does not matter w.r.t first several terms.
Then, you can see
$frac{1}{sqrt{n} -frac{2}{3}} sim frac{1}{sqrt{n}} = frac{1}{n^{1/2}}$
notice that for those power less than or equals 1,(here it's 1/2), it's a diverge sequence.
(You may refer to any analysis book for this result.)
answered Dec 5 '18 at 20:24
汪铈达汪铈达
111
answered Dec 5 '18 at 20:24
汪铈达汪铈达
111
answered Dec 5 '18 at 20:24
汪铈达汪铈达
111
answered Dec 5 '18 at 20:24
汪铈达汪铈达
111
111
add a comment |
add a comment |
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$begingroup$
looks good to me
$endgroup$
– gt6989b
Dec 5 '18 at 20:19