Check convergence /divergence of a series $sum_{n=1}^{infty} (-1)^{n+1}(n)^{frac{1}{4}}(frac{1}{sqrt{4n-1}}-...












1












$begingroup$



for $(a)$ and $(b)$ decide if the series converges , converges conditionally or diverge




$(a)$ $sum_{n=1}^{infty} (-1)^{n+1}(n)^{frac{1}{4}}(frac{1}{sqrt{4n-1}}- frac{1}{sqrt{4n}} )$



$(b) $ $sum_{n=1}^{infty} (-1)^{n+1}(n)^{frac{5}{4}}(frac{1}{sqrt{4n-1}}- frac{1}{sqrt{4n}} )$



i tried to use Leibniz test for $(a)$ and $(b)$ but failed.
i tried to do assymptotic tests but failed.



also i am not sure how to decide if the sequence monotonically decreasing






sequences-and-series limits convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    First series is absolutely convergent. Compare with $sum frac 1 {n^{5/4}}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 9:14












  • $begingroup$
    but i can't compare since its not positive series
    $endgroup$
    – Mather
    Jan 1 at 9:16










  • $begingroup$
    i know i tried with the absolute values but the lim is $ infty - infty $ how to calculate it i am trying lophital too
    $endgroup$
    – Mather
    Jan 1 at 9:26










  • $begingroup$
    devide by $frac{1}{ n^{ frac{5}{4} } }$ the lim is unclear
    $endgroup$
    – Mather
    Jan 1 at 9:29










  • $begingroup$
    For (a), Sum is equal to integral: $frac{int_0^{infty } frac{-1+, _1F_1left(frac{1}{2};frac{1}{4};frac{x}{4}right)}{left(1+e^xright) x^{3/4}} , dx}{2 Gamma left(frac{1}{4}right)}approx 0.0585087$
    $endgroup$
    – Mariusz Iwaniuk
    Jan 1 at 10:01


















1












$begingroup$



for $(a)$ and $(b)$ decide if the series converges , converges conditionally or diverge




$(a)$ $sum_{n=1}^{infty} (-1)^{n+1}(n)^{frac{1}{4}}(frac{1}{sqrt{4n-1}}- frac{1}{sqrt{4n}} )$



$(b) $ $sum_{n=1}^{infty} (-1)^{n+1}(n)^{frac{5}{4}}(frac{1}{sqrt{4n-1}}- frac{1}{sqrt{4n}} )$



i tried to use Leibniz test for $(a)$ and $(b)$ but failed.
i tried to do assymptotic tests but failed.



also i am not sure how to decide if the sequence monotonically decreasing






sequences-and-series limits convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    First series is absolutely convergent. Compare with $sum frac 1 {n^{5/4}}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 9:14












  • $begingroup$
    but i can't compare since its not positive series
    $endgroup$
    – Mather
    Jan 1 at 9:16










  • $begingroup$
    i know i tried with the absolute values but the lim is $ infty - infty $ how to calculate it i am trying lophital too
    $endgroup$
    – Mather
    Jan 1 at 9:26










  • $begingroup$
    devide by $frac{1}{ n^{ frac{5}{4} } }$ the lim is unclear
    $endgroup$
    – Mather
    Jan 1 at 9:29










  • $begingroup$
    For (a), Sum is equal to integral: $frac{int_0^{infty } frac{-1+, _1F_1left(frac{1}{2};frac{1}{4};frac{x}{4}right)}{left(1+e^xright) x^{3/4}} , dx}{2 Gamma left(frac{1}{4}right)}approx 0.0585087$
    $endgroup$
    – Mariusz Iwaniuk
    Jan 1 at 10:01
















1












1








1





$begingroup$



for $(a)$ and $(b)$ decide if the series converges , converges conditionally or diverge




$(a)$ $sum_{n=1}^{infty} (-1)^{n+1}(n)^{frac{1}{4}}(frac{1}{sqrt{4n-1}}- frac{1}{sqrt{4n}} )$



$(b) $ $sum_{n=1}^{infty} (-1)^{n+1}(n)^{frac{5}{4}}(frac{1}{sqrt{4n-1}}- frac{1}{sqrt{4n}} )$



i tried to use Leibniz test for $(a)$ and $(b)$ but failed.
i tried to do assymptotic tests but failed.



also i am not sure how to decide if the sequence monotonically decreasing






sequences-and-series limits convergence







share|cite|improve this question











$endgroup$





for $(a)$ and $(b)$ decide if the series converges , converges conditionally or diverge




$(a)$ $sum_{n=1}^{infty} (-1)^{n+1}(n)^{frac{1}{4}}(frac{1}{sqrt{4n-1}}- frac{1}{sqrt{4n}} )$



$(b) $ $sum_{n=1}^{infty} (-1)^{n+1}(n)^{frac{5}{4}}(frac{1}{sqrt{4n-1}}- frac{1}{sqrt{4n}} )$



i tried to use Leibniz test for $(a)$ and $(b)$ but failed.
i tried to do assymptotic tests but failed.



also i am not sure how to decide if the sequence monotonically decreasing






sequences-and-series limits convergence




sequences-and-series limits convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 9:55









user376343

3,9033829




3,9033829










asked Jan 1 at 9:06









Mather Mather

3418




3418












  • $begingroup$
    First series is absolutely convergent. Compare with $sum frac 1 {n^{5/4}}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 9:14












  • $begingroup$
    but i can't compare since its not positive series
    $endgroup$
    – Mather
    Jan 1 at 9:16










  • $begingroup$
    i know i tried with the absolute values but the lim is $ infty - infty $ how to calculate it i am trying lophital too
    $endgroup$
    – Mather
    Jan 1 at 9:26










  • $begingroup$
    devide by $frac{1}{ n^{ frac{5}{4} } }$ the lim is unclear
    $endgroup$
    – Mather
    Jan 1 at 9:29










  • $begingroup$
    For (a), Sum is equal to integral: $frac{int_0^{infty } frac{-1+, _1F_1left(frac{1}{2};frac{1}{4};frac{x}{4}right)}{left(1+e^xright) x^{3/4}} , dx}{2 Gamma left(frac{1}{4}right)}approx 0.0585087$
    $endgroup$
    – Mariusz Iwaniuk
    Jan 1 at 10:01




















  • $begingroup$
    First series is absolutely convergent. Compare with $sum frac 1 {n^{5/4}}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 1 at 9:14












  • $begingroup$
    but i can't compare since its not positive series
    $endgroup$
    – Mather
    Jan 1 at 9:16










  • $begingroup$
    i know i tried with the absolute values but the lim is $ infty - infty $ how to calculate it i am trying lophital too
    $endgroup$
    – Mather
    Jan 1 at 9:26










  • $begingroup$
    devide by $frac{1}{ n^{ frac{5}{4} } }$ the lim is unclear
    $endgroup$
    – Mather
    Jan 1 at 9:29










  • $begingroup$
    For (a), Sum is equal to integral: $frac{int_0^{infty } frac{-1+, _1F_1left(frac{1}{2};frac{1}{4};frac{x}{4}right)}{left(1+e^xright) x^{3/4}} , dx}{2 Gamma left(frac{1}{4}right)}approx 0.0585087$
    $endgroup$
    – Mariusz Iwaniuk
    Jan 1 at 10:01


















$begingroup$
First series is absolutely convergent. Compare with $sum frac 1 {n^{5/4}}$.
$endgroup$
– Kavi Rama Murthy
Jan 1 at 9:14






$begingroup$
First series is absolutely convergent. Compare with $sum frac 1 {n^{5/4}}$.
$endgroup$
– Kavi Rama Murthy
Jan 1 at 9:14














$begingroup$
but i can't compare since its not positive series
$endgroup$
– Mather
Jan 1 at 9:16




$begingroup$
but i can't compare since its not positive series
$endgroup$
– Mather
Jan 1 at 9:16












$begingroup$
i know i tried with the absolute values but the lim is $ infty - infty $ how to calculate it i am trying lophital too
$endgroup$
– Mather
Jan 1 at 9:26




$begingroup$
i know i tried with the absolute values but the lim is $ infty - infty $ how to calculate it i am trying lophital too
$endgroup$
– Mather
Jan 1 at 9:26












$begingroup$
devide by $frac{1}{ n^{ frac{5}{4} } }$ the lim is unclear
$endgroup$
– Mather
Jan 1 at 9:29




$begingroup$
devide by $frac{1}{ n^{ frac{5}{4} } }$ the lim is unclear
$endgroup$
– Mather
Jan 1 at 9:29












$begingroup$
For (a), Sum is equal to integral: $frac{int_0^{infty } frac{-1+, _1F_1left(frac{1}{2};frac{1}{4};frac{x}{4}right)}{left(1+e^xright) x^{3/4}} , dx}{2 Gamma left(frac{1}{4}right)}approx 0.0585087$
$endgroup$
– Mariusz Iwaniuk
Jan 1 at 10:01






$begingroup$
For (a), Sum is equal to integral: $frac{int_0^{infty } frac{-1+, _1F_1left(frac{1}{2};frac{1}{4};frac{x}{4}right)}{left(1+e^xright) x^{3/4}} , dx}{2 Gamma left(frac{1}{4}right)}approx 0.0585087$
$endgroup$
– Mariusz Iwaniuk
Jan 1 at 10:01












1 Answer
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$begingroup$

Answer to first series: $frac 1 {sqrt {(4n-1)}}- frac 1 {sqrt {4n}} =frac {sqrt {4n}-sqrt {4n-1}} {sqrt {4n-1} sqrt {4n}}$. Write this as $frac 1 {sqrt {4n-1} sqrt {4n}(sqrt {4n}+sqrt {4n-1})}$. Note that the denominator is of the order of $n^{3/2}$. Apply comparison test now (Note that $frac 3 2-frac 1 4=frac 5 4 >1)$. Hence the first series is absolutely convergent. A similar argument shows that the second series is not absolutely convergent. Try to prove its convergence using Alternating Series Test.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you sir , much clear now
    $endgroup$
    – Mather
    Jan 1 at 9:54











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Answer to first series: $frac 1 {sqrt {(4n-1)}}- frac 1 {sqrt {4n}} =frac {sqrt {4n}-sqrt {4n-1}} {sqrt {4n-1} sqrt {4n}}$. Write this as $frac 1 {sqrt {4n-1} sqrt {4n}(sqrt {4n}+sqrt {4n-1})}$. Note that the denominator is of the order of $n^{3/2}$. Apply comparison test now (Note that $frac 3 2-frac 1 4=frac 5 4 >1)$. Hence the first series is absolutely convergent. A similar argument shows that the second series is not absolutely convergent. Try to prove its convergence using Alternating Series Test.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you sir , much clear now
    $endgroup$
    – Mather
    Jan 1 at 9:54
















2












$begingroup$

Answer to first series: $frac 1 {sqrt {(4n-1)}}- frac 1 {sqrt {4n}} =frac {sqrt {4n}-sqrt {4n-1}} {sqrt {4n-1} sqrt {4n}}$. Write this as $frac 1 {sqrt {4n-1} sqrt {4n}(sqrt {4n}+sqrt {4n-1})}$. Note that the denominator is of the order of $n^{3/2}$. Apply comparison test now (Note that $frac 3 2-frac 1 4=frac 5 4 >1)$. Hence the first series is absolutely convergent. A similar argument shows that the second series is not absolutely convergent. Try to prove its convergence using Alternating Series Test.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you sir , much clear now
    $endgroup$
    – Mather
    Jan 1 at 9:54














2












2








2





$begingroup$

Answer to first series: $frac 1 {sqrt {(4n-1)}}- frac 1 {sqrt {4n}} =frac {sqrt {4n}-sqrt {4n-1}} {sqrt {4n-1} sqrt {4n}}$. Write this as $frac 1 {sqrt {4n-1} sqrt {4n}(sqrt {4n}+sqrt {4n-1})}$. Note that the denominator is of the order of $n^{3/2}$. Apply comparison test now (Note that $frac 3 2-frac 1 4=frac 5 4 >1)$. Hence the first series is absolutely convergent. A similar argument shows that the second series is not absolutely convergent. Try to prove its convergence using Alternating Series Test.






share|cite|improve this answer









$endgroup$



Answer to first series: $frac 1 {sqrt {(4n-1)}}- frac 1 {sqrt {4n}} =frac {sqrt {4n}-sqrt {4n-1}} {sqrt {4n-1} sqrt {4n}}$. Write this as $frac 1 {sqrt {4n-1} sqrt {4n}(sqrt {4n}+sqrt {4n-1})}$. Note that the denominator is of the order of $n^{3/2}$. Apply comparison test now (Note that $frac 3 2-frac 1 4=frac 5 4 >1)$. Hence the first series is absolutely convergent. A similar argument shows that the second series is not absolutely convergent. Try to prove its convergence using Alternating Series Test.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 1 at 9:35









Kavi Rama MurthyKavi Rama Murthy

63.9k42464




63.9k42464












  • $begingroup$
    thank you sir , much clear now
    $endgroup$
    – Mather
    Jan 1 at 9:54


















  • $begingroup$
    thank you sir , much clear now
    $endgroup$
    – Mather
    Jan 1 at 9:54
















$begingroup$
thank you sir , much clear now
$endgroup$
– Mather
Jan 1 at 9:54




$begingroup$
thank you sir , much clear now
$endgroup$
– Mather
Jan 1 at 9:54


















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