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

The truncated Fourier cosine series gives the better approximation over the whole closed interval. Its Fourier coefficients obey $|a_n|\le 4/(\pi n^2)$, so absolute convergence and the tail boundgive uniform convergence. The even periodic extension is continuous; its slope changes only cause corners.

By contrast $b_n=(-1{)}^{n+1}/n+O(n^{-2})$. The odd periodic extension has a jump at $\pi$, giving slower coefficient decay and the Gibbs phenomenon near that endpoint. Every truncated Fourier sine series satisfies $S_N(\pi )=0$, soIt therefore cannot converge uniformly to $f$ on $[0,\pi ]$, although it converges at every point of $[0,\pi )$. This comparison concerns approximation across the interval; it does not assert that the cosine truncation has smaller error at every individual interior point for every $N$.

The series $\sum u_n$ has absolute convergence if $\sum |u_n|$ is a convergent series, meaning its nonnegative partial sums have a finite limit. Absolute convergence implies convergence of the original real or complex series. The converse fails: the alternating harmonic series converges, while its absolute-value series is the divergent harmonic series.

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.

The comparison test for series states that if $0\le a_n\le b_n$ for all sufficiently large $n$, then convergence of $\sum b_n$ implies convergence of $\sum a_n$; conversely, divergence of $\sum a_n$ implies divergence of $\sum b_n$. A summable bound on $|a_n|$ proves absolute convergence of $\sum a_n$.

For the ratio test, suppose the terms are eventually nonzero and $L={\lim }_{n\to \infty }|a_{n+1}/a_n|$ exists. If $L<1$, the series converges absolutely. If $L>1$, including $L=\infty$, the terms fail to tend to zero and the series diverges. If $L=1$, the test is inconclusive. An eventually zero series converges separately. Convergence means that the sequence of partial sums has a finite limit; the terms tending to zero is only a necessary condition for a convergent series.

At $x=0$, the Fourier series equals $f(0)=0$. Isolating the first term givesChanging the sign of the remaining denominators provesThe absolute convergence justifies the evaluation and rearrangement.

The ratio test uses $L={\lim }_{n\to \infty }|a_{n+1}/a_n|$, when it exists for eventually nonzero terms: $L<1$ gives absolute convergence, $L>1$ gives divergence, and $L=1$ is inconclusive.

For $a_n=n/(2^n-1)$,Thus the first series converges absolutely. For the alternating second series,so it diverges. Alternating signs cannot rescue terms whose absolute values fail to tend to zero.

For a series with eventually nonzero terms $a_n$, suppose $r={\lim }_{n\to \infty }|a_{n+1}/a_n|$ exists. The ratio test says that $r<1$ implies absolute convergence, while $r>1$ implies divergence because the terms fail to tend to zero. At $r=1$ the test is inconclusive.

For absolute convergence, examine ${\sum }_{n\ge 1}|x-3{|}^n/n$. When $0<|x-3|<1$, the ratio of successive terms tends to $|x-3|<1$, so the ratio test proves convergence; at $x=3$ every term is zero. For $|x-3|>1$, the terms do not tend to zero.

At either boundary $|x-3|=1$, the absolute-value series is the divergent harmonic series. Thus for real $x$ the complete answer isThe original series at $x=2$ is conditionally convergent by the alternating-series criterion, but it is not absolutely convergent and is correctly excluded here.

Use the real Fourier series convention $f=a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$. The function is an even function, so $b_n=0$. For noninteger real $\mu$,For $n\ge 1$, the product-to-sum identity givesThus the Fourier series of a noninteger-frequency cosine isIts Fourier coefficients are $O(n^{-2})$, giving absolute convergence and uniform convergence. The periodic function is continuous at its joins because the two endpoint cosine values coincide, so the Dirichlet-Jordan convergence theorem gives this value at every point.

Putting $\mu =1/3$ givesEvaluate at $x=0$, where $f(0)=1$, and multiply by $\pi /\sqrt{3}$:Comparing with the required form gives $\boxed{A=3,B=3}$.

Insert $u=-1/x^2$ into the exponential series $e^u={\sum }_{n=0}^{\infty }{u}^n/n!$. For every fixed $x\ne 0$, this gives the expansion with absolute convergenceThis is a Laurent series in inverse powers of $x$, as requested. It is not a Maclaurin series about zero: that series has every coefficient zero and cannot equal $f$ on a punctured neighborhood of zero. There is no contradiction, since the inverse-power expansion does not apply at $x=0$.

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 $N\to \infty$.

The even-power formula reduces every summand to a scalar multiple of $I$:Both series have absolute convergence. Apply the logarithm expansion ${\sum }_{n\ge 1}{z}^n/n=-\log (1-z)$ for $|z|<1$ to obtainThus the off-diagonal entries vanish and both diagonal entries have the required value.

The integrand for $I_1$ is an odd function with absolute convergence, so $I_1=0$. For $I_2$, the recurrence gives $I_2=I_0/2$, where the supplied Gaussian integral is $I_0=\sqrt{\pi }$. HenceThe zero first moment is specific to the symmetric full-line domain.

The integral converges absolutely: near zero it behaves like $\log x$, and at infinity like $(\log x)/x^2$. Let its value be $J$. The integration by substitution $x=1/t$ givesThereforeAbsolute convergence makes this cancellation between the two ends legitimate.

Write the formal Maclaurin series as ${\sum }_{n\ge 0}{a}_n{x}^n$, where $a_n=f^{(n)}(0)/n!$. If the successive nonzero coefficient ratios have a limit, the ratio test givesso the series converges absolutely for $|x|<R$, whereIf that ratio does not exist, the Cauchy-Hadamard theorem gives the general formula $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. The radius of a Maclaurin series from successive derivatives concerns convergence of the series; equality to the original function additionally requires analyticity or a vanishing Taylor remainder. Mere differentiability, or even smoothness, does not ensure that equality.

For the specified logarithm,The absolute successive-term ratio tends to $|x|/2$, so the interval of absolute convergence is $(-2,2)$. At $x=2$ the nonconstant terms are the alternating harmonic series, which converges conditionally to $\ln 2$, giving the full sum $\ln 4$. At $x=-2$ all nonconstant terms are $-1/n$, so the series diverges to $-\infty$; the original logarithm is also undefined there. Thus the full real convergence interval is $(-2,2]$.

For $r\ne 1$, differentiate the finite geometric series and multiply by $r$:The required terms are $-n{r}^n$ with $r=-1/3$. The finite weighted geometric sum therefore givesSince $N/3^N\to 0$, the correction tends to zero, proving convergence and yieldingIn fact the ratio test also proves absolute convergence, since the ratio of consecutive absolute terms tends to $1/3$.

Let $a_n=\ln [1+(-1{)}^n/n]$ and $S_N={\sum }_{n=2}^N{a}_n$. Consecutive even and odd terms cancel exactly:ThusBoth parity subsequences have the same limit, so the full sequence of partial sums converges and the sum is zero. This check is necessary: cancellation in fixed pairs alone does not always prove convergence of the original series.

For absolute convergence, each pair instead contributesThe inequality follows from $\ln (1+u)={\int }_0^u(1+s{)}^{-1}ds\ge u/(1+u)$. The lower-bound series diverges by comparison with the harmonic series. Therefore the paired logarithmic series with zero sum has

The usual interior condition for the binomial series is $|3x|<1$, giving $-1/3<x<1/3$ as its open interval of convergence. In this particular square-root case the full series also converges absolutely at both endpoints, so the maximal real range on which its sum equals the original function isFor an endpoint check, write $\sqrt{1-u}=1-{\sum }_{n\ge 1}{b}_n{u}^n$, where $b_n=a_{n-1}-a_n>0$ and $a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\to 0$. Then $\sum b_n=1$ by telescoping, establishing absolute convergence at $u=\pm 1$. The square-root binomial series at its endpoints proves continuity of the sum there. The endpoint values are $\sqrt{2}$ at $x=-1/3$ and $0$ at $x=1/3$. Outside this range the power-series terms do not tend to zero. A three-term truncation remains an approximation, rather than an exact formula throughout the interval.

Both sums are geometric series. For first term $a$ and common ratio $r$ with $|r|<1$, the sum is $a/(1-r)$. Taking the initial term to be $2$ givesandBoth have absolute convergence because the series of absolute values has ratio $1/2$.