Why $sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = 1/(1-k)$? The Next CEO of Stack Overflowmaybe this sum have approximation $sum_k=0^nbinomnk^3approxfrac2pisqrt3ncdot 8^n,ntoinfty$How prove binomial cofficients $sum_k=0^[fracn3](-1)^kbinomn+1kbinom2n-3kn=sum_k=[fracn2]^nbinomn+1kbinomkn-k$How find the sum $2sum_k=1^inftysum_i=0^2k-1fracbinom2kicdot B_icdot(m-1)^2k-i2k(2k-1)m^2k-1$How to prove that $sum_k=0^infty binomxx-kcdotbinomxk-x = 1$?prove that $(frac16)^4cdotlim_nrightarrowinftysum_i=4^nbinomi-13(frac56)^i-4=1$Evaluate$sum_n=0^inftybinom3nnx^n$$sum_i=1^ni^k=sum_i=1^kfracc_k+1-ik!i!n^i$. What is known about $c_k_k=0^infty$?$lim_limitssigma_A, sigma_B to infty e^-sigma_A-sigma_B sum_k=0^infty fracsigma_A^kk!cdotfracsigma_B^k+1(k+1)!$Proof for binomial theorem $ (1+x)^a = sum_k=0^infty P_k(a)x^k $Help understanding proof of the following statement $E(Y) = sum_i = 1^infty P(Y geq k)$
How do I go from 300 unfinished/half written blog posts, to published posts?
Why does C# sound extremely flat when saxophone is tuned to G?
Is it safe to use c_str() on a temporary string?
Can a caster that cast Polymorph on themselves stop concentrating at any point even if their Int is low?
How can I quit an app using Terminal?
Is it okay to store user locations?
How do spells that require an ability check vs. the caster's spell save DC work?
Visit to the USA with ESTA approved before trip to Iran
If the heap is initialized for security, then why is the stack uninitialized?
How do I get the green key off the shelf in the Dobby level of Lego Harry Potter 2?
If I blow insulation everywhere in my attic except the door trap, will heat escape through it?
What's the point of interval inversion?
Shade part of a Venn diagram
Why were Madagascar and New Zealand discovered so late?
Why do professional authors make "consistency" mistakes? And how to avoid them?
How did people program for Consoles with multiple CPUs?
What do "high sea" and "carry" mean in this sentence?
Inappropriate reference requests from Journal reviewers
How to Reset Passwords on Multiple Websites Easily?
Can I equip Skullclamp on a creature I am sacrificing?
Term for the "extreme-extension" version of a straw man fallacy?
Which organization defines CJK Unified Ideographs?
What does this shorthand mean?
Was a professor correct to chastise me for writing "Prof. X" rather than "Professor X"?
Why $sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = 1/(1-k)$?
The Next CEO of Stack Overflowmaybe this sum have approximation $sum_k=0^nbinomnk^3approxfrac2pisqrt3ncdot 8^n,ntoinfty$How prove binomial cofficients $sum_k=0^[fracn3](-1)^kbinomn+1kbinom2n-3kn=sum_k=[fracn2]^nbinomn+1kbinomkn-k$How find the sum $2sum_k=1^inftysum_i=0^2k-1fracbinom2kicdot B_icdot(m-1)^2k-i2k(2k-1)m^2k-1$How to prove that $sum_k=0^infty binomxx-kcdotbinomxk-x = 1$?prove that $(frac16)^4cdotlim_nrightarrowinftysum_i=4^nbinomi-13(frac56)^i-4=1$Evaluate$sum_n=0^inftybinom3nnx^n$$sum_i=1^ni^k=sum_i=1^kfracc_k+1-ik!i!n^i$. What is known about $c_k_k=0^infty$?$lim_limitssigma_A, sigma_B to infty e^-sigma_A-sigma_B sum_k=0^infty fracsigma_A^kk!cdotfracsigma_B^k+1(k+1)!$Proof for binomial theorem $ (1+x)^a = sum_k=0^infty P_k(a)x^k $Help understanding proof of the following statement $E(Y) = sum_i = 1^infty P(Y geq k)$
$begingroup$
$$
sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = frac11-k
$$
I take this equal than I do some problem of probabiliti teory. I think that it will be prove by indukcia, but I can't.
probability binomial-coefficients
$endgroup$
add a comment |
$begingroup$
$$
sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = frac11-k
$$
I take this equal than I do some problem of probabiliti teory. I think that it will be prove by indukcia, but I can't.
probability binomial-coefficients
$endgroup$
$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40
add a comment |
$begingroup$
$$
sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = frac11-k
$$
I take this equal than I do some problem of probabiliti teory. I think that it will be prove by indukcia, but I can't.
probability binomial-coefficients
$endgroup$
$$
sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = frac11-k
$$
I take this equal than I do some problem of probabiliti teory. I think that it will be prove by indukcia, but I can't.
probability binomial-coefficients
probability binomial-coefficients
edited Mar 18 at 12:56
Arthur
120k7121206
120k7121206
asked Mar 18 at 12:53
Andrey KomisarovAndrey Komisarov
726
726
$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40
add a comment |
$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40
$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
it helps to write $i=j+l$, then divide out $(1-k)^j$ to restate the claim as $sum_lge 0binomj+ljk^l=(1-k)^-j-1$. This is just a special case of the binomial theorem, since the falling Pochhammer symbol $(-j-1)_l=binomj+lj(-1)^l$.
$endgroup$
add a comment |
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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3152745%2fwhy-sum-i-j-infty-fracii-j-cdot-j-cdot-ki-j-cdot-1-kj-1%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
it helps to write $i=j+l$, then divide out $(1-k)^j$ to restate the claim as $sum_lge 0binomj+ljk^l=(1-k)^-j-1$. This is just a special case of the binomial theorem, since the falling Pochhammer symbol $(-j-1)_l=binomj+lj(-1)^l$.
$endgroup$
add a comment |
$begingroup$
it helps to write $i=j+l$, then divide out $(1-k)^j$ to restate the claim as $sum_lge 0binomj+ljk^l=(1-k)^-j-1$. This is just a special case of the binomial theorem, since the falling Pochhammer symbol $(-j-1)_l=binomj+lj(-1)^l$.
$endgroup$
add a comment |
$begingroup$
it helps to write $i=j+l$, then divide out $(1-k)^j$ to restate the claim as $sum_lge 0binomj+ljk^l=(1-k)^-j-1$. This is just a special case of the binomial theorem, since the falling Pochhammer symbol $(-j-1)_l=binomj+lj(-1)^l$.
$endgroup$
it helps to write $i=j+l$, then divide out $(1-k)^j$ to restate the claim as $sum_lge 0binomj+ljk^l=(1-k)^-j-1$. This is just a special case of the binomial theorem, since the falling Pochhammer symbol $(-j-1)_l=binomj+lj(-1)^l$.
answered Mar 18 at 13:00
J.G.J.G.
32.3k23250
32.3k23250
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3152745%2fwhy-sum-i-j-infty-fracii-j-cdot-j-cdot-ki-j-cdot-1-kj-1%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40