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By the mean value theorem, $|f(x)-f(y)|\le A|x-y|$ for all real $x,y$. Thus $f$ is a contraction mapping on the complete metric space $\mathbb{R}$. Starting from arbitrary $x_1$, define $x_{n+1}=f(x_n)$. Induction gives $|x_{n+1}-x_n|\le A^{n-1}|x_2-x_1|$. For $m>n$, the triangle inequality and a geometric series yieldHence $(x_n)$ is a Cauchy sequence and completeness of the real numbers supplies a limit $x$. The Lipschitz continuity of $f$ implies $f(x)=\lim f(x_n)=\lim x_{n+1}=x$. Thus the limit is a fixed point, independently of the initial value. If $y$ were another, $|x-y|\le A|x-y|$ would force $x=y$. This proves the contraction mapping theorem in this setting, with error estimate

Writing $q=1-p$, the probability generating function is a geometric series:The displayed restriction ensures absolute convergence.

Summing the joint probability mass function over $y$ with the geometric series gives the marginal probability mass functionDivide the joint probability mass function by this positive marginal probability mass function to obtain the conditional distribution:It is the geometric distribution counting failures before the first success, with success probability $1-q^{10}$. For the other nonnegative integers $x\ge 10$, $\mathbb{P}(X=x)=0$; the elementary conditional probability given $X=x$ is undefined and supplies no additional case.

Integrate the geometric series for $1/(1+x)$ from zero to $x$. The first three nonzero terms areThe resulting Taylor series converges for $|x|<1$.

With replacement, successive draws are independent, with red probability $2/7$ and green probability $5/7$. The first red on draw $n$ requires $n-1$ greens followed by one red. HenceThis is a geometric distribution. Its total probability is one because the geometric series sums to $(2/7)/(1-5/7)=1$.

The summands $u_n=n^4/2^n$ are positive. Apply the ratio test:ThereforeFor a direct comparison interpretation, the ratio is eventually at most any fixed number between $1/2$ and $1$, such as $3/4$. The tail is then bounded by a convergent geometric series. Thus the exponential denominator wins over the fourth-power numerator strongly enough to make the whole series summable.

This is a geometric series with ratio $r=2\alpha /({\alpha }^2+1)$. Since ${\alpha }^2+1>0$, the condition $|r|<1$ is equivalent toIt holds for every real $\alpha$ except $\alpha =\pm 1$. At $\alpha =1$ the terms are all one; at $\alpha =-1$ they alternate between $-1$ and $1$. Neither sequence of terms tends to zero. ThusAll the convergent cases have absolute convergence, including $\alpha =0$, when every term is zero.

Use the geometric series $(1+u{)}^{-1}=1-u+u^2+O(u^3)$ with $u=\sin x=x+O(x^3)$. The constant term counts as the first nonzero term, so the requested expansion is

This is a two-turn renewal game. A full pair of throws has probability $(1/2{)}^2=1/4$ of passing twice and returning to the original state. The probability that $A$ wins before that restart is $5/12$. Thus first-step analysis gives $p=5/12+p/4$, orEquivalently sum the geometric series $(5/12){\sum }_{j\ge 0}(1/4{)}^j$. The probability of surviving infinitely many pairs is zero, so the eventual winner is well defined with probability one.

The geometric series has first term $1$ and common ratio $1/3$, whose absolute value is less than one. Therefore

Normalization of a probability distribution and the geometric series give $1=A/(1-\rho )$, so $A=1-\rho$. For a nonnegative integer $n$, the strict tail starts at $n+1$:For nonnegative integers $n>m$, the event $\{N>n\}$ is contained in $\{N>m\}$. ThereforeThis is the memorylessness of the geometric distribution with the zero-based counting convention. Keeping strict versus non-strict inequalities explicit is essential: $N-m-1$, conditional on $N>m$, has the original distribution of $N$.

For $|t|<1$, the geometric series gives $(1+t{)}^{-1}={\sum }_{n\ge 0}(-t{)}^n$. Its uniform convergence on every closed interval strictly inside $(-1,1)$ permits termwise integration from zero to $x$. ThereforeSubtracting cancels every even power and doubles every odd power, giving the Taylor series

This geometric series has first term one and ratio $-1/4$, whose absolute value is less than one. Therefore

This geometric series has first term $1$ and ratio $0.999$, which lies strictly inside the unit interval. ThereforeThe decimal is treated as the exact value $999/1000$.

Since $(1/2{)}^{3k}=(1/8{)}^k$, this is a geometric series of first term $1$ and common ratio $1/8$. Its sum is

This geometric series has first term $2^{-4}=1/16$ and common ratio $1/2$, whose absolute value is less than one. Consequently

Apply the ratio test to the infinite series whose partial sums are $S_N$. Assuming the terms are eventually nonzero and the limit exists, letIf $\ell <1$, the series has absolute convergence, by eventual comparison with a decaying geometric series. If $\ell >1$, including an infinite limit, the terms fail to tend to zero and the series diverges. If $\ell =1$, the test is inconclusive: both $\sum 1/k$ and $\sum 1/k^2$ have that ratio limit, but only the second converges. A finite partial sum itself requires no convergence test; the issue is the limit as $N\to \infty$.

For $p\le 0$, the terms $k^{-p}$ do not tend to zero, so the series diverges. For $p>0$, group the positive terms into the dyadic blocks $2^j\le k<2^{j+1}$. Each block contains $2^j$ terms, and monotonicity givesWhen $p>1$, the upper bounds form a convergent geometric series with ratio $2^{1-p}<1$, proving convergence. When $0<p\le 1$, the lower bounds do not tend to zero, so infinitely many blocks each make a nonvanishing contribution and the sum diverges. Thus the p-series criterion isThis dyadic comparison is the mechanism of the Cauchy condensation test. In particular $p=1$ is the divergent harmonic series.

The derivative of the inverse tangent is $(1+x^2{)}^{-1}$. Expand it as a geometric series and integrate term by term near zero, using $\arctan 0=0$:

Index the stated geometric progression by $x_i=(1+ϵ{)}^i$, $i=0,1,\dots$. Then $\Delta x_i=ϵ(1+ϵ{)}^i$, and its left-endpoint contribution isHence the approximation is a geometric series withSince $ϵ>0$ and $\alpha >0$, the ratio satisfies $0<q<1$ and the series converges. The integrand is decreasing, so the left-endpoint sum is an upper bound for the improper integral.