Convergence of $sum_n=1^infty fracln^k(n) n^a$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite













Show when $displaystyle sum_n=1^infty fracln^k(n) n^a$ is convergent.




Tried using convergence tests. I tried to calculate
$$lim: lim_nto infty fracfracln^k(n) n^afrac1n^a$$
but that doesn't help me.
Need some advises, thanks






calculus sequences-and-series analysis







share|cite|improve this question























  • You should clarify the range for $a$ and $k$, otherwise we need to consider some cases.
    – gimusi
    Sep 1 at 7:48










  • Possible duplicate of Convergence of $sumlimits_n=2^infty frac1n^alpha ln^beta (n) $ for nonnegative $alpha$ and $beta$
    – Nosrati
    Sep 1 at 10:06














up vote
1
down vote

favorite













Show when $displaystyle sum_n=1^infty fracln^k(n) n^a$ is convergent.




Tried using convergence tests. I tried to calculate
$$lim: lim_nto infty fracfracln^k(n) n^afrac1n^a$$
but that doesn't help me.
Need some advises, thanks






calculus sequences-and-series analysis







share|cite|improve this question























  • You should clarify the range for $a$ and $k$, otherwise we need to consider some cases.
    – gimusi
    Sep 1 at 7:48










  • Possible duplicate of Convergence of $sumlimits_n=2^infty frac1n^alpha ln^beta (n) $ for nonnegative $alpha$ and $beta$
    – Nosrati
    Sep 1 at 10:06












up vote
1
down vote

favorite









up vote
1
down vote

favorite












Show when $displaystyle sum_n=1^infty fracln^k(n) n^a$ is convergent.




Tried using convergence tests. I tried to calculate
$$lim: lim_nto infty fracfracln^k(n) n^afrac1n^a$$
but that doesn't help me.
Need some advises, thanks






calculus sequences-and-series analysis







share|cite|improve this question
















Show when $displaystyle sum_n=1^infty fracln^k(n) n^a$ is convergent.




Tried using convergence tests. I tried to calculate
$$lim: lim_nto infty fracfracln^k(n) n^afrac1n^a$$
but that doesn't help me.
Need some advises, thanks






calculus sequences-and-series analysis




calculus sequences-and-series analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 1 at 7:34









Robert Z

85.6k1055123




85.6k1055123










asked Sep 1 at 7:20









Maor Rocky

876




876











  • You should clarify the range for $a$ and $k$, otherwise we need to consider some cases.
    – gimusi
    Sep 1 at 7:48










  • Possible duplicate of Convergence of $sumlimits_n=2^infty frac1n^alpha ln^beta (n) $ for nonnegative $alpha$ and $beta$
    – Nosrati
    Sep 1 at 10:06
















  • You should clarify the range for $a$ and $k$, otherwise we need to consider some cases.
    – gimusi
    Sep 1 at 7:48










  • Possible duplicate of Convergence of $sumlimits_n=2^infty frac1n^alpha ln^beta (n) $ for nonnegative $alpha$ and $beta$
    – Nosrati
    Sep 1 at 10:06















You should clarify the range for $a$ and $k$, otherwise we need to consider some cases.
– gimusi
Sep 1 at 7:48




You should clarify the range for $a$ and $k$, otherwise we need to consider some cases.
– gimusi
Sep 1 at 7:48












Possible duplicate of Convergence of $sumlimits_n=2^infty frac1n^alpha ln^beta (n) $ for nonnegative $alpha$ and $beta$
– Nosrati
Sep 1 at 10:06




Possible duplicate of Convergence of $sumlimits_n=2^infty frac1n^alpha ln^beta (n) $ for nonnegative $alpha$ and $beta$
– Nosrati
Sep 1 at 10:06










1 Answer
1






active

oldest

votes

















up vote
2
down vote













HINT



We should declare the range for $a$ and $k$, otherwise we need to consider the following cases:



  • For $a>1$ take $b=frac1+a2$ and apply limit comparison test with
    $sum frac1n^b$.

  • For $0le a<1$ and $kge0$ use comparison test with $sum
    frac1n^a$.

  • For $0le a<1$ and $k<0$ refer to Cauchy condensation test.

  • For $a<0$ we have that $a_n not to 0$.

Note also that when $k<0$ we need to start from $n=2$ in order to have a well defined expression.






share|cite|improve this answer






















  • Concrete and satisfying+
    – mrs
    Sep 1 at 8:15






  • 1




    @ResidentDementor Thanks for your kind appreciation!Maybe I should add that for the case $k<0$ we need to start from $n=2$. Bye!
    – gimusi
    Sep 1 at 8:19










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2901457%2fconvergence-of-sum-n-1-infty-frac-lnkn-na%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













HINT



We should declare the range for $a$ and $k$, otherwise we need to consider the following cases:



  • For $a>1$ take $b=frac1+a2$ and apply limit comparison test with
    $sum frac1n^b$.

  • For $0le a<1$ and $kge0$ use comparison test with $sum
    frac1n^a$.

  • For $0le a<1$ and $k<0$ refer to Cauchy condensation test.

  • For $a<0$ we have that $a_n not to 0$.

Note also that when $k<0$ we need to start from $n=2$ in order to have a well defined expression.






share|cite|improve this answer






















  • Concrete and satisfying+
    – mrs
    Sep 1 at 8:15






  • 1




    @ResidentDementor Thanks for your kind appreciation!Maybe I should add that for the case $k<0$ we need to start from $n=2$. Bye!
    – gimusi
    Sep 1 at 8:19














up vote
2
down vote













HINT



We should declare the range for $a$ and $k$, otherwise we need to consider the following cases:



  • For $a>1$ take $b=frac1+a2$ and apply limit comparison test with
    $sum frac1n^b$.

  • For $0le a<1$ and $kge0$ use comparison test with $sum
    frac1n^a$.

  • For $0le a<1$ and $k<0$ refer to Cauchy condensation test.

  • For $a<0$ we have that $a_n not to 0$.

Note also that when $k<0$ we need to start from $n=2$ in order to have a well defined expression.






share|cite|improve this answer






















  • Concrete and satisfying+
    – mrs
    Sep 1 at 8:15






  • 1




    @ResidentDementor Thanks for your kind appreciation!Maybe I should add that for the case $k<0$ we need to start from $n=2$. Bye!
    – gimusi
    Sep 1 at 8:19












up vote
2
down vote










up vote
2
down vote









HINT



We should declare the range for $a$ and $k$, otherwise we need to consider the following cases:



  • For $a>1$ take $b=frac1+a2$ and apply limit comparison test with
    $sum frac1n^b$.

  • For $0le a<1$ and $kge0$ use comparison test with $sum
    frac1n^a$.

  • For $0le a<1$ and $k<0$ refer to Cauchy condensation test.

  • For $a<0$ we have that $a_n not to 0$.

Note also that when $k<0$ we need to start from $n=2$ in order to have a well defined expression.






share|cite|improve this answer














HINT



We should declare the range for $a$ and $k$, otherwise we need to consider the following cases:



  • For $a>1$ take $b=frac1+a2$ and apply limit comparison test with
    $sum frac1n^b$.

  • For $0le a<1$ and $kge0$ use comparison test with $sum
    frac1n^a$.

  • For $0le a<1$ and $k<0$ refer to Cauchy condensation test.

  • For $a<0$ we have that $a_n not to 0$.

Note also that when $k<0$ we need to start from $n=2$ in order to have a well defined expression.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 1 at 8:30

























answered Sep 1 at 7:23









gimusi

71.9k73888




71.9k73888











  • Concrete and satisfying+
    – mrs
    Sep 1 at 8:15






  • 1




    @ResidentDementor Thanks for your kind appreciation!Maybe I should add that for the case $k<0$ we need to start from $n=2$. Bye!
    – gimusi
    Sep 1 at 8:19
















  • Concrete and satisfying+
    – mrs
    Sep 1 at 8:15






  • 1




    @ResidentDementor Thanks for your kind appreciation!Maybe I should add that for the case $k<0$ we need to start from $n=2$. Bye!
    – gimusi
    Sep 1 at 8:19















Concrete and satisfying+
– mrs
Sep 1 at 8:15




Concrete and satisfying+
– mrs
Sep 1 at 8:15




1




1




@ResidentDementor Thanks for your kind appreciation!Maybe I should add that for the case $k<0$ we need to start from $n=2$. Bye!
– gimusi
Sep 1 at 8:19




@ResidentDementor Thanks for your kind appreciation!Maybe I should add that for the case $k<0$ we need to start from $n=2$. Bye!
– gimusi
Sep 1 at 8:19

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2901457%2fconvergence-of-sum-n-1-infty-frac-lnkn-na%23new-answer', 'question_page');

);

Post as a guest
































































-

這個網誌中的熱門文章

How to combine Bézier curves to a surface?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
閱讀完整內容

Mutual Information Always Non-negative

圖片
Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
閱讀完整內容