Any good upper bounds for $\sum_{i = 0}^{m} 2^i {k+i \choose i}$? I am trying to solve a recursion (I faced it when I wanted to analyze the running time of an algorithm precisely). I reduced it to compute a "good" upper bound for $\sum_{i = 0}^{m}2^i {k + i \choose i}$. We have these assumptions that $1 < m \leq Ck$, for some $C\in \mathbb{N}$.
I have tried many approximations but the bounds are not what I want. They are very loose. On the other hand, it doesn't look like a very sophisticated summation but I have so much difficulty solving it!
Does anyone have any ideas?
 A: $$\sum_{i=0}^m2^i\binom{k+i}{i}=\sum_{i=0}^m2^i\binom{k+i}{k}\le \sum_{i=0}^m2^i\cdot \binom{k+m}{k}=\binom{k+m}{k}\cdot (2^{m+1}-1)$$
Since $\sum_{i=0}^m2^i\binom{k+i}{i}$ is at least as big as the last term, $2^m\binom{k+m}{k}$, this bound is at most a factor of $2$ away from correct value.
A: Not a full answer and not rigorous, but somebody more knowledgeable may be able to justify the asymptotics below.
If we set a specific value for $k$, we can compute easily enough a formula of the form:
$$\sum_{i=0}^{m}{2^i{k+i \choose i}} = P_k(m)2^m+(-1)^k$$
where $P_k(m)=a_{k}m^k+a_{k-1}m^{k-1}+\ldots+a_0$, and experimentally (to be proved) it looks like, for $k \ge 2$:
$$a_k = \frac{2}{k!}$$
I have retrieved some of the $P_k(m)$ with the help of OEIS:
$$P_1(m) = m - 1$$
$$P_2(m) = m^2+m+2$$
$$P_3(m) = \frac{1}{3}(m^3+3m^2+8m)$$
$$P_4(m) = \frac{1}{12}(m^4+6m^3+23m^2+18m+24)$$
$$P_5(m) = \frac{1}{60}(m^5+10m^4+55m^3+110m^2+184m)$$
and, in general, a conjecture, for $m$ big and $k \ge 2$:
$$\sum_{i=0}^{m}{2^i{k+i \choose i}} \approx \frac{m^{k}2^{m+1}}{k!}$$
A: Define $\{a_m\}$ by
$$a_m = \sum_{i=0}^m 2^i \binom{k+i}{i}$$
Then we can show
$$a_m = \frac{2^{m+1} m^k}{k!} \left( 1 + O \left(\frac{1}{m} \right) \right)$$
by application of Note IV.26, page 256, in Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick.
If you should happen to consult the book, be aware that Note IV.26 contains a typographical error: the factor $\alpha_1^{-n+r}$ should be $\alpha_1^{-n-r}$.
Note IV.26, corrected and slightly paraphrased, reads:
Let $f(z)$ be a rational generating function with a single dominant
pole at $z=\alpha_1$ with multiplicity $r$.  Then  $$f_n =
 \frac{C}{(r-1)!} \alpha_1^{-n-r} n^{r-1} \left( 1 + O \left(
 \frac{1}{n} \right) \right)$$ with $$C = \lim_{z \to \alpha_1}
 (z-\alpha_1)^r f(z)$$
[end of Note]
A "dominant pole" is a pole nearest the origin.
The generating function for $a_m$ is
$$f(z) = \frac{1}{(1-z)(1-2z)^{k+1}}$$
which has a dominant pole at $z=1/2$ of multiplicity $r=k+1$.
Applying Note IV.26 we find $C= (1/2)^k$, yielding the asymptotic relation stated above.
A: I'm pretty sure there is no closed form for this sum and it is far from simple.  This sum looks like one involved in finding the expected value of the length of the longest run of successes of a bernoulli random variable.  See for instance this question, unfortunately the it's pretty opaque and it's been a while since I was trying to do that sum.
Here's a way of converting the problem to a derivative problem, although I don't know how useful it will be.  Recall $\binom{k+i}{i}=\binom{k+i}{k}=[x^k y^i](x+y)^{k+i}$, where $[m](p)$ denotes the coefficient on the term $m$ in the polynomial/series $p$.  So now we want to find
$$\sum_{i=0}^m 2^i[x^k y^i](x+y)^{k+i}.$$
Notice if we take the $k$th derivative of $(x+y)^{k+i}$ with respect to $x$ at $x=0$ and $y=1$, we get the coefficient we want times $k!$.  Thus we can write the sum as
$$\sum_{i=0}^m\frac{2^i}{k!}\frac{d^k}{dx^k}(x+1)^{k+i}|_{x=0}$$
Now a lot of this stuff can be pulled out of the sum, giving us
$$\frac{1}{k!}\frac{d^k}{dx^k}\left((x+1)^k\sum_{i=0}^m 2^i(x+1)^i\right)|_{x=0}$$
The inside is now a geometric series so we can find its sum
$$\frac{1}{k!}\frac{d^k}{dx^k}\left((x+1)^k\frac{(2(x+1))^{m+1}-1}{2(x+1)-1}\right)|_{x=0}$$
$$\frac{1}{k!}\frac{d^k}{dx^k}\left(\frac{2^{m+1}(x+1)^{m+k+1}-(x+1)^k}{2x-1}\right)|_{x=0}$$
It may be possible to find an explicit form for this derivative or at least a recurrence relation, although I couldn't.
However, if $k$ is fixed you can find a closed form: the binomial coefficient $\binom{k+i}{k}$ is a degree $k$ polynomial, and $\sum_n b^n P(n)$ can be computed for any base $b$ and polynomial $P$ using a trick.  Take $\frac{d^k}{db^k}\sum_n b^n$ for $k$ ranging from 1 to the degree of $P$.  The derivatives of the summands form a basis for polynomials of degree $k$, and the derivatives of the corresponding sums let us calculate the sum for $b^n$ times any polynomial.
As far as an upper bound, a VERY rough one can be gotten from the approximation $\binom{mn}{n}\approx\frac{m^{m(n-1)+1}}{(m-1)^{(m-1)(n-1)}\sqrt{n}}$.  Because $i\le m\le Ck$, we can rewrite the sum as $$\sum_{i=0}^m\frac{2^i(C+1)^{Ck-C+k}}{C^{Ck-C}\sqrt{k}}=\frac{(2^{m+1}-1)(C+1)^{Ck-C+k}}{C^{Ck-C}\sqrt{k}}.$$
Notice this is equivalent to using the upper bound from the other answer and then applying the approximation I mentioned, so this will also be over by a factor of about 2.  Depending on what you need (ie if you just want big O), a factor of 2 may not matter to you.
A: First step is that  we can actually give to the sum a closed expression in terms of the Hypergeometric function
In fact can rewrite the sum as follows (allow me to change a bit your notation)
$$
\eqalign{
  & S(n,m,x) = \sum\limits_{i = 0}^m {\left( \matrix{ n + i \cr i \cr}  \right)x^{\,\,i} }
  = \sum\limits_{i = 0}^m {\left( { - 1} \right)^{\,\,i} \left( \matrix{ - n - 1 \cr i \cr}  \right)x^{\,\,i} }  = \quad \quad (1)  \cr 
  &  = \sum\limits_{0\, \le \,i\, \le \,m} {\left( \matrix{ - n - 1 \cr i \cr}  \right)\left( { - x} \right)^{\,\,i} }
  = \left( { - x} \right)^{\,\,m} \sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)}
 {\left( \matrix{ - n - 1 \cr  m - k \cr}  \right)\left( { - x} \right)^{\, - k} }  = \quad \quad (2)  \cr 
  &  = x^{\,\,m} \sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)}
 {\left( \matrix{n + m - k \cr m - k \cr}  \right)x^{\, - k} }
  = x^{\,\,m} \sum\limits_{0\, \le \,k\, \le \,m}
 {{{\left( {n + m - k} \right)!} \over {n!\left( {m - k} \right)!}}x^{\, - k} }  = \quad \quad (3)  \cr 
  &  = x^{\,\,m} {{\left( {n + m} \right)!} \over {m!n!}}\sum\limits_{0\, \le \,k\, \le \,m}
 {{{\left( {n + m - k} \right)!m!} \over {\left( {n + m} \right)!\left( {m - k} \right)!}}x^{\, - k} }  = \quad \quad (4)  \cr 
  &  = x^{\,\,m} \left( \matrix{ n + m \cr m \cr}  \right)\sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)}
 {{{\left( \matrix{m \cr k \cr}  \right)} \over {\left( \matrix{n + m \cr k \cr}  \right)}}x^{\, - k} }  = \quad \quad (5)  \cr 
  &  = x^{\,\,m} \left( \matrix{n + m \cr  m \cr}  \right)\sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)}
 {{{m^{\,\underline {\,k\,} } } \over {\left( {n + m} \right)^{\,\underline {\,k\,} } }}x^{\, - k} }  = \quad \quad (6)  \cr 
  &  = x^{\,\,m} \left( \matrix{n + m \cr m \cr}  \right)\sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)}
 {{{\left( { - m} \right)^{\,\overline {\,k\,} } }
 \over {\left( { - n - m} \right)^{\,\overline {\,k\,} } }}x^{\, - k} }  = \quad \quad (7)  \cr 
  &  = x^{\,\,m} \left( \matrix{  n + m \cr m \cr}  \right)
 {}_2F_{\,1} \left( {\left. {\matrix{ {1,\; - m}  \cr { - \left( {n + m} \right)}  \cr } \;} \right|\;{1 \over x}} \right)
 \quad \quad (8) \cr} 
$$
where $a^{\,\underline {\,k\,} } ,\quad a^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial
and the various steps are:
-1) upper negation;

*

*


*invert the summation index;



*


*binomial expressed as factorials;



*


*extraction of the leading binomial component;



*


*, 6), 7) rewriting of the binomial fraction



*


*conversion to Hypergeometric.
Note the upper bound of the sum being indicated in brackets when it is implicit in the binomial or falling factorial.



For what concerns the establishing of an upper bound, it seems that you are referring
to the case in which $n << m$.
If it is so, we can use one of the expression above, eg. the (6). The coefficient in the sum can be rewritten in various ways
$$
\eqalign{
  & {{m^{\,\underline {\,k\,} } } \over {\left( {m + n} \right)^{\,\underline {\,k\,} } }}\,
 = {{\left( {m + n - k} \right)^{\,\underline {\,n\,} } } \over {\left( {m + n} \right)^{\,\underline {\,n\,} } }}
 = {{\left( {m + 1 - k} \right)^{\,\overline {\,n\,} } } \over {\left( {m + 1} \right)^{\,\overline {\,n\,} } }} =   \cr 
  &  = {{\Gamma \left( {m + 1 + n - k} \right)\Gamma \left( {m + 1} \right)}
 \over {\Gamma \left( {m + 1 - k} \right)\Gamma \left( {m + 1 + n} \right)}} =   \cr 
  &  = \prod\limits_{j = 0}^{n - 1} {{{m + 1 + j - k} \over {m + 1 + j}}}
  = \prod\limits_{j = 0}^{n - 1} {\left( {1 - {k \over {m + 1 + j}}} \right)}  \cr} 
$$
and from these we can proceed with various approximations, depending on the required acurracy.
For instance
$$
{{m^{\,\underline {\,k\,} } } \over {\left( {m + n} \right)^{\,\underline {\,k\,} } }}
 = \prod\limits_{j = 0}^{n - 1} {\left( {1 - {k \over {m + 1 + j}}} \right)}
  < \mathop  \approx \limits_{n <  < m} \left( {1 - {k \over {m + \left( {n + 1} \right)/2}}} \right)^n 
$$
and then proceeding to take the log and simplify (with the accuracy needed), e.g.
$$
 \approx \exp \left( { - {{nk} \over {m + \left( {n + 1} \right)/2}}} \right)
$$
which is suitable to be summed in $k$.
Or you can start from the expression through the Gamma function and apply the Stirling approximation, ...
