Since $X^2=X(X-1)+X$, the second moment is $\mathbb{E}[X^2]=G^{\prime \prime }(1)+G^{\prime }(1)$. Subtracting the squared mean gives

Under the extra hypothesis that $G(s)$ exists for every real $s$, its power series has infinite radius of convergence. Termwise differentiation of a power series is therefore valid to every finite order. For $k\ge 1$,At $s=1$ this is the expected value of the falling factorial of $X$:In particular $\boxed{G^{\prime }(1)=\mathbb{E}[X]}$. The terms with $j<k$ vanish because their falling factorials contain a zero factor. These factorial moments are finite under the stated hypothesis.

For a nonnegative integer-valued random variable, its probability generating function isThe convention $s^0=1$, including at $s=0$, gives $G(0)=p(0)$. The power series has absolute convergence for every real $|s|\le 1$, since ${\sum }_{j}p(j)|s{|}^j\le 1$. More generally its real domain as a finite expected value is $\{s\in \mathbb{R}:{\sum }_{j}p(j)|s{|}^j<\infty \}$, which may extend beyond $[-1,1]$. No extension beyond that guaranteed interval is automatic.

Each $X_i$ has the same probability distribution as $-X_i$, and independence makes $S_m$ and $-S_m$ identically distributed. Thus their two strict tail probabilities are equal:Apply Chebyshev's inequality using the mean and variance from part (i):The equal-tail argument is essential: applying Chebyshev's inequality directly to the absolute tail alone gives the weaker bound $8/(3m)$.

Each random variable has mean zero and second moment $2/3$, so its variance is $2/3$. For $S_m={\sum }_{i=1}^m{X}_i$, linearity of expectation and variance additivity for independent random variables give

At an absorbing state, the remaining absorption time is zero, so $e_0=e_n=0$. Multiplying the linear recurrence relation by three givesThe second finite difference of $k^2$ is two. A particular solution is therefore $-3k^2/2$, while every solution of the corresponding homogeneous linear recurrence relation has the form $A+Bk$. The boundary conditions give $A=0$ and $B=3n/2$, henceThis is the mean absorption time of a lazy symmetric random walk with holding probability $1/3$.

First the expected value of the absorption time is finite. From any interior state, consecutive leftward moves lead to absorption within at most $n$ steps, with probability at least $p=3^{-n}$. By independence of the subsequent steps, the probability of surviving $jn$ steps is at most $(1-p{)}^j$. Summing this geometric series bounds the expected value of the absorption time by $n/p$.

Now apply first-step analysis. Starting from $k$ in the interior, one step is always used. Conditional on that step, the remaining expected values are $e_{k+1}$, $e_k$, and $e_{k-1}$, each with probability $1/3$. The law of total expectation therefore givesThe term $1$ remains even when the particle stays at $k$, because a step has elapsed. Rearranging gives

For an integrable random variable $X$, the law of total expectation for the partition of a sample space isParts of zero probability contribute zero and need no definition of $\mathbb{E}[X\mid B_i]$. To see the identity, write $X={\sum }_{i}X{1}_{B_i}$ and use linearity of expectation; integrability justifies the countable sum since ${\sum }_i\mathbb{E}[|X|1_{B_i}]=\mathbb{E}[|X|]<\infty$.

The events $B_1,B_2,\dots$ form a partition of a sample space $\Omega$ when they are pairwise disjoint sets and cover the whole sample space:Thus exactly one of these events occurs on every outcome. Under the usual set partition convention the parts are nonempty; empty parts can simply be discarded.

Choose $h_n=\min \{1/n,(b-c)/2\}>0$. Then $h_n\to 0$ and $[c,c+h_n]\subset (a,b)$. The mean value theorem states that a continuous function on a closed interval, differentiable in its interior, has an interior derivative equal to its endpoint difference quotient. It gives $x_n\in (c,c+h_n)$ withHere $x_n\ne c$, $x_n\to c$, and the right-hand side tends to $f^{\prime }(c)$ by differentiability at $c$. If $L={\lim }_{x\to c}{f}^{\prime }(x)$ exists, the same values $f^{\prime }(x_n)$ must tend to $L$. Uniqueness of the limit of a sequence givesThus a derivative cannot have a removable discontinuity at a point where the original function is differentiable. The proof also excludes an infinite limit of a function there, since the displayed difference quotients have a finite limit of a sequence.

Fix $d$ strictly between $f^{\prime }(a)$ and $f^{\prime }(b)$, and set $F(x)=f(x)-dx$. The function $F$ is continuous on $[a,b]$: interior differentiability gives interior continuity, and the finite one-sided derivatives give continuity at the endpoints. The extreme value theorem therefore gives a point $c$ where $F$ attains its minimum.

Since $F_{+}^{\prime }(a)=f^{\prime }(a)-d<0$, for sufficiently small $h>0$ we have $F(a+h)<F(a)$, so $a$ cannot be a minimizing point. Since $F_{-}^{\prime }(b)=f^{\prime }(b)-d>0$, the quotient $(F(b-h)-F(b))/(-h)$ is positive for sufficiently small $h>0$, giving $F(b-h)<F(b)$. Thus $b$ cannot be a minimizing point either, and $c\in (a,b)$.

Fermat's theorem on stationary points states that a differentiable function at an interior local minimum or local maximum has derivative zero there. Applying it to $F$ givesThis proves the relevant case of Darboux's theorem; the argument uses continuity of $f$, with no assumption of continuity of $f^{\prime }$.

On $(0,1)$, the product rule and chain rule give $f^{\prime }(x)=2x\sin (1/x)-\cos (1/x)$. On $(-1,0)$, the function is constant and has derivative zero. At zero,Its absolute value is at most $|h|$, so the squeeze theorem gives $f^{\prime }(0)=0$. ThereforeFor $x_n=1/(2\pi n)$, we have $x_n\to 0$ but $f^{\prime }(x_n)=-1$ for every $n$. This differs from $f^{\prime }(0)=0$, so $f^{\prime }$ is not continuous at zero. Indeed, $y_n=1/((2n+1)\pi )$ gives $f^{\prime }(y_n)=1$, so the right-hand limit of a function does not exist. Thus $f$ is differentiable everywhere on $(-1,1)$, while $f^{\prime }$ is discontinuous at zero.

A function $f:(a,b)\to \mathbb{R}$ is differentiable on $(a,b)$ when, at each $c\in (a,b)$, its difference quotient has a finite limit of a function:Here $h\ne 0$ and $c+h\in (a,b)$.

For sufficiently small nonzero $h$, the difference quotient has absolute value at most $|f^{\prime }(c)|+1$, because it tends to $f^{\prime }(c)$. ConsequentlyThis is continuity at $c$. Since $c$ was arbitrary, differentiability implies continuity on $(a,b)$.

For real-valued functions, the Uniform Cauchy criterion applied to partial sums says that $\sum u_n$ has uniform convergence exactly whenSufficiency uses completeness of the real numbers.

Let $M$ be a common bound for the uniformly bounded family of functions $(g_n)$; increase it if necessary so that $M>0$. The assumed uniform absolute convergence of $\sum f_n$ gives $N$ such thatThe triangle inequality now yieldsThe Uniform Cauchy criterion therefore proves uniform convergence of $\sum f_n{g}_n$. The same estimate with $|f_n{g}_n|$ proves uniform absolute convergence as well.

Choose the unbounded function $\boxed{g(x)=1/x}$ on $E=(0,1/2]$. It is unbounded because $g(1/k)=k$ for all integers $k\ge 2$. For each $x\in E$, the geometric series gives the sum and its remainder after degree $N$:Since $0<x\le 1/2$,This proves uniform convergence. Although the first term $1/x$ and the sum are unbounded, the late tails have a bound tending to zero independently of $x$. Uniform convergence concerns those tails and does not require the individual functions to be bounded.

Let $S_N={\sum }_{n=1}^N{f}_n$ and let $s$ be its uniform limit. Choose a finite $B\ge 0$ with $|g(x)|\le B$ for all $x\in E$. The partial sums of the multiplied series are $g{S}_N$, andFor any $\epsilon >0$, uniform convergence supplies $N_0$ making $|S_N(x)-s(x)|<\epsilon /(B+1)$ for all $N\ge N_0$ and $x\in E$. The displayed estimate then gives an error less than $\epsilon$. Thus

Take $0\le r<R$. At the point $r$, the power series has absolute convergence, so ${\sum }_{n=0}^{\infty }|a_n|r^n<\infty$. On the whole interval $[-r,r]$,The Weierstrass M-test with $M_n=|a_n|r^n$ proves uniform convergence on $[-r,r]$. It also gives the explicit tail estimateFor $r=0$ the interval consists only of zero, and all positive-degree terms vanish. If $R=\infty$, the same argument applies to every finite $r$; if $R=0$, there is no $r\ge 0$ with $r<R$.

The Weierstrass M-test states: if functions $f_n:E\to \mathbb{R}$ or $\mathbb{C}$ satisfythen $\sum f_n$ has uniform convergence and absolute convergence on $E$. In fact $\sum |f_n|$ also has uniform convergence, since every tail is bounded by the corresponding tail of $\sum M_n$:

Write the partial sums as $S_N(x)={\sum }_{n=1}^N{f}_n(x)$. The series has uniform convergence to $s:E\to \mathbb{R}$ ifThe same threshold $N_0$ must control every point of $E$. Equivalently, the supremum error ${\sup }_{x\in E}|S_N(x)-s(x)|$ tends to zero, for nonempty $E$.

TakeBoth functions have values in $[1,\infty )$. The reverse triangle inequality gives $|g(x)-g(y)|\le |x-y|$, so $g$ is Lipschitz continuous and hence has uniform continuity.

For $x_n=n$ and $y_n=n+1/n$, the distances satisfy $|x_n-y_n|=1/n\to 0$, whereasIn particular tolerance $\epsilon =1$ defeats every proposed $\delta$: choose $n$ with $1/n<\delta$. Thus $f$ does not have uniform continuity on $\mathbb{R}$.