Determine if the infinite series converge or diverge
The series is as follows:
begin{equation}
sum {frac{n-2}{nsqrt{4n+2}}}
end{equation}
Which test suits the best for determing the convergence/divergence of this series?
I tried the Ratio test,the Limit comparison with ${frac{1}{n}}$ and with ${frac{1}{n^2}}$, I can't seem to find a "bigger" series that converges to use the Direct Comparison test and all of them with no result. I can't use the root test.
Any hint on how to get started on this one is appreciated.
Thanks in prior
calculus sequences-and-series
add a comment |
The series is as follows:
begin{equation}
sum {frac{n-2}{nsqrt{4n+2}}}
end{equation}
Which test suits the best for determing the convergence/divergence of this series?
I tried the Ratio test,the Limit comparison with ${frac{1}{n}}$ and with ${frac{1}{n^2}}$, I can't seem to find a "bigger" series that converges to use the Direct Comparison test and all of them with no result. I can't use the root test.
Any hint on how to get started on this one is appreciated.
Thanks in prior
calculus sequences-and-series
1
Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
– Alex
Dec 3 '18 at 20:10
Thank you very much Alex. I didn't see this one coming.
– Allorja
Dec 3 '18 at 20:23
add a comment |
The series is as follows:
begin{equation}
sum {frac{n-2}{nsqrt{4n+2}}}
end{equation}
Which test suits the best for determing the convergence/divergence of this series?
I tried the Ratio test,the Limit comparison with ${frac{1}{n}}$ and with ${frac{1}{n^2}}$, I can't seem to find a "bigger" series that converges to use the Direct Comparison test and all of them with no result. I can't use the root test.
Any hint on how to get started on this one is appreciated.
Thanks in prior
calculus sequences-and-series
The series is as follows:
begin{equation}
sum {frac{n-2}{nsqrt{4n+2}}}
end{equation}
Which test suits the best for determing the convergence/divergence of this series?
I tried the Ratio test,the Limit comparison with ${frac{1}{n}}$ and with ${frac{1}{n^2}}$, I can't seem to find a "bigger" series that converges to use the Direct Comparison test and all of them with no result. I can't use the root test.
Any hint on how to get started on this one is appreciated.
Thanks in prior
calculus sequences-and-series
calculus sequences-and-series
asked Dec 3 '18 at 20:06
Allorja
479
479
1
Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
– Alex
Dec 3 '18 at 20:10
Thank you very much Alex. I didn't see this one coming.
– Allorja
Dec 3 '18 at 20:23
add a comment |
1
Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
– Alex
Dec 3 '18 at 20:10
Thank you very much Alex. I didn't see this one coming.
– Allorja
Dec 3 '18 at 20:23
1
1
Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
– Alex
Dec 3 '18 at 20:10
Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
– Alex
Dec 3 '18 at 20:10
Thank you very much Alex. I didn't see this one coming.
– Allorja
Dec 3 '18 at 20:23
Thank you very much Alex. I didn't see this one coming.
– Allorja
Dec 3 '18 at 20:23
add a comment |
3 Answers
3
active
oldest
votes
Here's a way to think about this: Constants don't mean anything compared to the growth of $n^p$ for any $p>0.$ Thus instead of
$$sum frac{n-2}{nsqrt {4n+2}},$$
think about
$$sum frac{n}{nsqrt {n}} = sum frac{1}{sqrt {n}}.$$
Since the last series diverges, your guess should be "divergence" for the original series. That's the way calculus experts think about this. You still have to make sure your guess is correct, but instead of wandering around in the darkness, you have a good guide for the "first guess," and that is often the hardest part.
This just made my calculus easier for the rest of the semester. Thank you sir!
– Allorja
Dec 4 '18 at 15:56
add a comment |
For $nge 4$, $n-2gefrac{n}{2}$ while $4n+2lefrac{9n}{2}$, so $frac{n-2}{nsqrt{4n+2}}gefrac{1}{sqrt{18n}}$. This confirms divergence.
+1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
– Allorja
Dec 3 '18 at 20:25
add a comment |
In my opinion the simpler way is to note that
$$frac{n-2}{nsqrt{4n+2}}simfrac{n}{nsqrt{4n}}=frac1{2sqrt n}$$
and then refer to limit comparison test with $sum frac1{sqrt n}$.
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here's a way to think about this: Constants don't mean anything compared to the growth of $n^p$ for any $p>0.$ Thus instead of
$$sum frac{n-2}{nsqrt {4n+2}},$$
think about
$$sum frac{n}{nsqrt {n}} = sum frac{1}{sqrt {n}}.$$
Since the last series diverges, your guess should be "divergence" for the original series. That's the way calculus experts think about this. You still have to make sure your guess is correct, but instead of wandering around in the darkness, you have a good guide for the "first guess," and that is often the hardest part.
This just made my calculus easier for the rest of the semester. Thank you sir!
– Allorja
Dec 4 '18 at 15:56
add a comment |
Here's a way to think about this: Constants don't mean anything compared to the growth of $n^p$ for any $p>0.$ Thus instead of
$$sum frac{n-2}{nsqrt {4n+2}},$$
think about
$$sum frac{n}{nsqrt {n}} = sum frac{1}{sqrt {n}}.$$
Since the last series diverges, your guess should be "divergence" for the original series. That's the way calculus experts think about this. You still have to make sure your guess is correct, but instead of wandering around in the darkness, you have a good guide for the "first guess," and that is often the hardest part.
This just made my calculus easier for the rest of the semester. Thank you sir!
– Allorja
Dec 4 '18 at 15:56
add a comment |
Here's a way to think about this: Constants don't mean anything compared to the growth of $n^p$ for any $p>0.$ Thus instead of
$$sum frac{n-2}{nsqrt {4n+2}},$$
think about
$$sum frac{n}{nsqrt {n}} = sum frac{1}{sqrt {n}}.$$
Since the last series diverges, your guess should be "divergence" for the original series. That's the way calculus experts think about this. You still have to make sure your guess is correct, but instead of wandering around in the darkness, you have a good guide for the "first guess," and that is often the hardest part.
Here's a way to think about this: Constants don't mean anything compared to the growth of $n^p$ for any $p>0.$ Thus instead of
$$sum frac{n-2}{nsqrt {4n+2}},$$
think about
$$sum frac{n}{nsqrt {n}} = sum frac{1}{sqrt {n}}.$$
Since the last series diverges, your guess should be "divergence" for the original series. That's the way calculus experts think about this. You still have to make sure your guess is correct, but instead of wandering around in the darkness, you have a good guide for the "first guess," and that is often the hardest part.
answered Dec 4 '18 at 0:40
zhw.
71.7k43075
71.7k43075
This just made my calculus easier for the rest of the semester. Thank you sir!
– Allorja
Dec 4 '18 at 15:56
add a comment |
This just made my calculus easier for the rest of the semester. Thank you sir!
– Allorja
Dec 4 '18 at 15:56
This just made my calculus easier for the rest of the semester. Thank you sir!
– Allorja
Dec 4 '18 at 15:56
This just made my calculus easier for the rest of the semester. Thank you sir!
– Allorja
Dec 4 '18 at 15:56
add a comment |
For $nge 4$, $n-2gefrac{n}{2}$ while $4n+2lefrac{9n}{2}$, so $frac{n-2}{nsqrt{4n+2}}gefrac{1}{sqrt{18n}}$. This confirms divergence.
+1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
– Allorja
Dec 3 '18 at 20:25
add a comment |
For $nge 4$, $n-2gefrac{n}{2}$ while $4n+2lefrac{9n}{2}$, so $frac{n-2}{nsqrt{4n+2}}gefrac{1}{sqrt{18n}}$. This confirms divergence.
+1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
– Allorja
Dec 3 '18 at 20:25
add a comment |
For $nge 4$, $n-2gefrac{n}{2}$ while $4n+2lefrac{9n}{2}$, so $frac{n-2}{nsqrt{4n+2}}gefrac{1}{sqrt{18n}}$. This confirms divergence.
For $nge 4$, $n-2gefrac{n}{2}$ while $4n+2lefrac{9n}{2}$, so $frac{n-2}{nsqrt{4n+2}}gefrac{1}{sqrt{18n}}$. This confirms divergence.
answered Dec 3 '18 at 20:10
J.G.
23k22137
23k22137
+1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
– Allorja
Dec 3 '18 at 20:25
add a comment |
+1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
– Allorja
Dec 3 '18 at 20:25
+1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
– Allorja
Dec 3 '18 at 20:25
+1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
– Allorja
Dec 3 '18 at 20:25
add a comment |
In my opinion the simpler way is to note that
$$frac{n-2}{nsqrt{4n+2}}simfrac{n}{nsqrt{4n}}=frac1{2sqrt n}$$
and then refer to limit comparison test with $sum frac1{sqrt n}$.
add a comment |
In my opinion the simpler way is to note that
$$frac{n-2}{nsqrt{4n+2}}simfrac{n}{nsqrt{4n}}=frac1{2sqrt n}$$
and then refer to limit comparison test with $sum frac1{sqrt n}$.
add a comment |
In my opinion the simpler way is to note that
$$frac{n-2}{nsqrt{4n+2}}simfrac{n}{nsqrt{4n}}=frac1{2sqrt n}$$
and then refer to limit comparison test with $sum frac1{sqrt n}$.
In my opinion the simpler way is to note that
$$frac{n-2}{nsqrt{4n+2}}simfrac{n}{nsqrt{4n}}=frac1{2sqrt n}$$
and then refer to limit comparison test with $sum frac1{sqrt n}$.
answered Dec 3 '18 at 20:25
gimusi
1
1
add a comment |
add a comment |
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1
Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
– Alex
Dec 3 '18 at 20:10
Thank you very much Alex. I didn't see this one coming.
– Allorja
Dec 3 '18 at 20:23