$sum_{n=1}^{infty} frac{(1/2) + (-1)^{n}}{n}$ converges or diverges?












1












$begingroup$


How to check if the series $$sum_{n=1}^{infty} frac{(1/2) + (-1)^{n}}{n}$$ converges or diverges?



When $n$ is odd, series is $sum frac{-1}{2n}$



When $n$ is even, series is $sum frac{3}{2n}$



This series is similar to the series
$$sum frac{-1}{2(2n-1)} + frac{3}{2(2n)}$$



$$= sum frac{8n-6}{8n(2n-1)}$$
Which is clearly divergent.
So, the given series is divergent.



Is this method right?
Please, suggest if there is some easier way.






sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your method is not correct. You distinguish $n$ even and $n$ odd for the partial sum not for the general term.
    $endgroup$
    – hamam_Abdallah
    Dec 26 '18 at 13:17


















1












$begingroup$


How to check if the series $$sum_{n=1}^{infty} frac{(1/2) + (-1)^{n}}{n}$$ converges or diverges?



When $n$ is odd, series is $sum frac{-1}{2n}$



When $n$ is even, series is $sum frac{3}{2n}$



This series is similar to the series
$$sum frac{-1}{2(2n-1)} + frac{3}{2(2n)}$$



$$= sum frac{8n-6}{8n(2n-1)}$$
Which is clearly divergent.
So, the given series is divergent.



Is this method right?
Please, suggest if there is some easier way.






sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your method is not correct. You distinguish $n$ even and $n$ odd for the partial sum not for the general term.
    $endgroup$
    – hamam_Abdallah
    Dec 26 '18 at 13:17
















1












1








1


0



$begingroup$


How to check if the series $$sum_{n=1}^{infty} frac{(1/2) + (-1)^{n}}{n}$$ converges or diverges?



When $n$ is odd, series is $sum frac{-1}{2n}$



When $n$ is even, series is $sum frac{3}{2n}$



This series is similar to the series
$$sum frac{-1}{2(2n-1)} + frac{3}{2(2n)}$$



$$= sum frac{8n-6}{8n(2n-1)}$$
Which is clearly divergent.
So, the given series is divergent.



Is this method right?
Please, suggest if there is some easier way.






sequences-and-series







share|cite|improve this question









$endgroup$




How to check if the series $$sum_{n=1}^{infty} frac{(1/2) + (-1)^{n}}{n}$$ converges or diverges?



When $n$ is odd, series is $sum frac{-1}{2n}$



When $n$ is even, series is $sum frac{3}{2n}$



This series is similar to the series
$$sum frac{-1}{2(2n-1)} + frac{3}{2(2n)}$$



$$= sum frac{8n-6}{8n(2n-1)}$$
Which is clearly divergent.
So, the given series is divergent.



Is this method right?
Please, suggest if there is some easier way.






sequences-and-series




sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 26 '18 at 13:10









MathsaddictMathsaddict

3619




3619












  • $begingroup$
    Your method is not correct. You distinguish $n$ even and $n$ odd for the partial sum not for the general term.
    $endgroup$
    – hamam_Abdallah
    Dec 26 '18 at 13:17




















  • $begingroup$
    Your method is not correct. You distinguish $n$ even and $n$ odd for the partial sum not for the general term.
    $endgroup$
    – hamam_Abdallah
    Dec 26 '18 at 13:17


















$begingroup$
Your method is not correct. You distinguish $n$ even and $n$ odd for the partial sum not for the general term.
$endgroup$
– hamam_Abdallah
Dec 26 '18 at 13:17






$begingroup$
Your method is not correct. You distinguish $n$ even and $n$ odd for the partial sum not for the general term.
$endgroup$
– hamam_Abdallah
Dec 26 '18 at 13:17












2 Answers
2






active

oldest

votes


















2












$begingroup$

hint



$$sum frac{(-1)^n}{n}$$



is convergent by alternate criteria.
$$frac 12sum frac 1n$$
is a divergent Riemann series.



The sum of a convergent and a divergent series is a Divergent one.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The sum should diverge. Right?
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:24










  • $begingroup$
    @Mathsaddict Yes of course. but Div +Div could converge.
    $endgroup$
    – hamam_Abdallah
    Dec 26 '18 at 13:25



















3












$begingroup$

The idea is correct, but not correctly expressed. Asserting that the given series converges is equivalent to the assertion that the sequence$$left(sum_{n=1}^Nfrac{frac12+(-1)^n}nright)_{Ninmathbb N}$$converges. If it does, then the sequence$$left(sum_{n=1}^{2N}frac{frac12+(-1)^n}nright)_{Ninmathbb N}$$converges too. But it follows from your computations that it doesn't.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    From my computations, sequence of partiat sum diverges, and so does the series.
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:29










  • $begingroup$
    You write as if that's different from what I wrote.
    $endgroup$
    – José Carlos Santos
    Dec 26 '18 at 13:41










  • $begingroup$
    I thought you are pointing towards some mistake in my calculations, since you wrote 'but' in the last line. Does your answer support my method to check if the series converges?
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:52













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

hint



$$sum frac{(-1)^n}{n}$$



is convergent by alternate criteria.
$$frac 12sum frac 1n$$
is a divergent Riemann series.



The sum of a convergent and a divergent series is a Divergent one.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The sum should diverge. Right?
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:24










  • $begingroup$
    @Mathsaddict Yes of course. but Div +Div could converge.
    $endgroup$
    – hamam_Abdallah
    Dec 26 '18 at 13:25
















2












$begingroup$

hint



$$sum frac{(-1)^n}{n}$$



is convergent by alternate criteria.
$$frac 12sum frac 1n$$
is a divergent Riemann series.



The sum of a convergent and a divergent series is a Divergent one.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The sum should diverge. Right?
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:24










  • $begingroup$
    @Mathsaddict Yes of course. but Div +Div could converge.
    $endgroup$
    – hamam_Abdallah
    Dec 26 '18 at 13:25














2












2








2





$begingroup$

hint



$$sum frac{(-1)^n}{n}$$



is convergent by alternate criteria.
$$frac 12sum frac 1n$$
is a divergent Riemann series.



The sum of a convergent and a divergent series is a Divergent one.






share|cite|improve this answer











$endgroup$



hint



$$sum frac{(-1)^n}{n}$$



is convergent by alternate criteria.
$$frac 12sum frac 1n$$
is a divergent Riemann series.



The sum of a convergent and a divergent series is a Divergent one.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 26 '18 at 13:42

























answered Dec 26 '18 at 13:14









hamam_Abdallahhamam_Abdallah

38.1k21634




38.1k21634












  • $begingroup$
    The sum should diverge. Right?
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:24










  • $begingroup$
    @Mathsaddict Yes of course. but Div +Div could converge.
    $endgroup$
    – hamam_Abdallah
    Dec 26 '18 at 13:25


















  • $begingroup$
    The sum should diverge. Right?
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:24










  • $begingroup$
    @Mathsaddict Yes of course. but Div +Div could converge.
    $endgroup$
    – hamam_Abdallah
    Dec 26 '18 at 13:25
















$begingroup$
The sum should diverge. Right?
$endgroup$
– Mathsaddict
Dec 26 '18 at 13:24




$begingroup$
The sum should diverge. Right?
$endgroup$
– Mathsaddict
Dec 26 '18 at 13:24












$begingroup$
@Mathsaddict Yes of course. but Div +Div could converge.
$endgroup$
– hamam_Abdallah
Dec 26 '18 at 13:25




$begingroup$
@Mathsaddict Yes of course. but Div +Div could converge.
$endgroup$
– hamam_Abdallah
Dec 26 '18 at 13:25











3












$begingroup$

The idea is correct, but not correctly expressed. Asserting that the given series converges is equivalent to the assertion that the sequence$$left(sum_{n=1}^Nfrac{frac12+(-1)^n}nright)_{Ninmathbb N}$$converges. If it does, then the sequence$$left(sum_{n=1}^{2N}frac{frac12+(-1)^n}nright)_{Ninmathbb N}$$converges too. But it follows from your computations that it doesn't.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    From my computations, sequence of partiat sum diverges, and so does the series.
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:29










  • $begingroup$
    You write as if that's different from what I wrote.
    $endgroup$
    – José Carlos Santos
    Dec 26 '18 at 13:41










  • $begingroup$
    I thought you are pointing towards some mistake in my calculations, since you wrote 'but' in the last line. Does your answer support my method to check if the series converges?
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:52


















3












$begingroup$

The idea is correct, but not correctly expressed. Asserting that the given series converges is equivalent to the assertion that the sequence$$left(sum_{n=1}^Nfrac{frac12+(-1)^n}nright)_{Ninmathbb N}$$converges. If it does, then the sequence$$left(sum_{n=1}^{2N}frac{frac12+(-1)^n}nright)_{Ninmathbb N}$$converges too. But it follows from your computations that it doesn't.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    From my computations, sequence of partiat sum diverges, and so does the series.
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:29










  • $begingroup$
    You write as if that's different from what I wrote.
    $endgroup$
    – José Carlos Santos
    Dec 26 '18 at 13:41










  • $begingroup$
    I thought you are pointing towards some mistake in my calculations, since you wrote 'but' in the last line. Does your answer support my method to check if the series converges?
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:52
















3












3








3





$begingroup$

The idea is correct, but not correctly expressed. Asserting that the given series converges is equivalent to the assertion that the sequence$$left(sum_{n=1}^Nfrac{frac12+(-1)^n}nright)_{Ninmathbb N}$$converges. If it does, then the sequence$$left(sum_{n=1}^{2N}frac{frac12+(-1)^n}nright)_{Ninmathbb N}$$converges too. But it follows from your computations that it doesn't.






share|cite|improve this answer









$endgroup$



The idea is correct, but not correctly expressed. Asserting that the given series converges is equivalent to the assertion that the sequence$$left(sum_{n=1}^Nfrac{frac12+(-1)^n}nright)_{Ninmathbb N}$$converges. If it does, then the sequence$$left(sum_{n=1}^{2N}frac{frac12+(-1)^n}nright)_{Ninmathbb N}$$converges too. But it follows from your computations that it doesn't.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 26 '18 at 13:16









José Carlos SantosJosé Carlos Santos

161k22127232




161k22127232












  • $begingroup$
    From my computations, sequence of partiat sum diverges, and so does the series.
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:29










  • $begingroup$
    You write as if that's different from what I wrote.
    $endgroup$
    – José Carlos Santos
    Dec 26 '18 at 13:41










  • $begingroup$
    I thought you are pointing towards some mistake in my calculations, since you wrote 'but' in the last line. Does your answer support my method to check if the series converges?
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:52




















  • $begingroup$
    From my computations, sequence of partiat sum diverges, and so does the series.
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:29










  • $begingroup$
    You write as if that's different from what I wrote.
    $endgroup$
    – José Carlos Santos
    Dec 26 '18 at 13:41










  • $begingroup$
    I thought you are pointing towards some mistake in my calculations, since you wrote 'but' in the last line. Does your answer support my method to check if the series converges?
    $endgroup$
    – Mathsaddict
    Dec 26 '18 at 13:52


















$begingroup$
From my computations, sequence of partiat sum diverges, and so does the series.
$endgroup$
– Mathsaddict
Dec 26 '18 at 13:29




$begingroup$
From my computations, sequence of partiat sum diverges, and so does the series.
$endgroup$
– Mathsaddict
Dec 26 '18 at 13:29












$begingroup$
You write as if that's different from what I wrote.
$endgroup$
– José Carlos Santos
Dec 26 '18 at 13:41




$begingroup$
You write as if that's different from what I wrote.
$endgroup$
– José Carlos Santos
Dec 26 '18 at 13:41












$begingroup$
I thought you are pointing towards some mistake in my calculations, since you wrote 'but' in the last line. Does your answer support my method to check if the series converges?
$endgroup$
– Mathsaddict
Dec 26 '18 at 13:52






$begingroup$
I thought you are pointing towards some mistake in my calculations, since you wrote 'but' in the last line. Does your answer support my method to check if the series converges?
$endgroup$
– Mathsaddict
Dec 26 '18 at 13:52




















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