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Write $q=35/36$. Alice wins on her $(n+1)$st turn if $n$ complete rounds fail and then she succeeds. These disjoint events give the geometric seriesConsequentlyThe probability of never obtaining a winner is ${\lim }_{n\to \infty }{q}^{2n}=0$. The small advantage over $1/2$ comes from taking the first turn.

The Maclaurin series give $e^{-x^3}=1-x^3+O(x^6)$ and $\cosh x=1+x^2/2+x^4/24+O(x^6)$. Inverting the latter using the geometric series givesMultiplying the two expansions yieldsThe remainder is $O(x^5)$ because the product already generates an $x^5$ term from $(-x^3)(-x^2/2)$.

The $i$th screen distance is $x_0/2^i$, starting at $i=1$. The previous expected value therefore scales to $E[Y_i]=x_0\ln 3/(\pi 2^i)$. To justify the infinite linearity of expectation, note that $|\tan {\theta }_i|\le \sqrt{3}$ throughout the restricted angular interval. ThusThis deterministic integrable bound justifies the expectation of an absolutely summable random series, even without assuming independent firing angles. Summing the geometric series gives

The initial term at $n=-1$ is three, and the common ratio is $1/3$. The convergent geometric series therefore sums to

Write ${ϵ}_k=E_0(k+\frac{1}{2})$ and use Lagrange multipliers $\alpha ,\beta$ for the particle-number and energy constraints. The stationary equations areThus $n_k=e^{-\alpha }{e}^{-\beta {ϵ}_k}$. Finite normalization over infinitely many levels requires $\beta >0$. Put $q=e^{-\beta E_0}$, so $0<q<1$. The geometric series givesThereforeFor the energy constraint, use $n_k=N(1-q)q^k$ and $\sum k{q}^k=q/(1-q{)}^2$, obtained by differentiating the geometric series. This yieldsRecognizing the hyperbolic cotangent givesThe strict positivity of all occupations requires $E>N{E}_0/2$; the paper's positivity assumptions alone do not guarantee this feasibility condition. For admissible energy the multiplier is uniquely $\beta =E_0^{-1}\ln [(2E+N{E}_0)/(2E-N{E}_0)]>0$. Equality $E=N{E}_0/2$ is only the limiting state with all occupation in level zero, outside the stipulated strictly positive family; lower energy is impossible.

For rigor beyond formal multiplier equations, $p_k=n_k/N$ has fixed mean $E/(N{E}_0)-1/2$. The objective differs from $N$ times its information entropy only by the fixed constant $N-N\ln N$. The maximum entropy distribution on the nonnegative integers is uniquely geometric at that mean, confirming the stationary family and its maximum character.

Multiply the Maclaurin series $\ln (1+x^2)=x^2-x^4/2+O(x^6)$ and the geometric series $(1+x^2{)}^{-1}=1-x^2+O(x^4)$. The first two nonzero terms are

Set $h=x+1$, the displacement from the expansion point. The numerator has Taylor series $\sinh h=h+h^3/6+O(h^5)$, while the reciprocal denominator has geometric series $(1+h{)}^{-1}=1-h+h^2-h^3+h^4+O(h^5)$. Multiplication givesThe local power series converges for $|x+1|<1$, the distance to the denominator's zero at $x=-2$.

For each fixed $n$, the inner geometric series sums to $2^{-n}/(1-1/2)=2^{1-n}$. Summing this second geometric series givesAll terms are nonnegative, so there is no conditional-rearrangement issue; the iterated sum is finite and the double series converges. Equivalently, collecting equal values of $m$ gives ${\sum }_{m=0}^{\infty }(m+1)2^{-m}=4$.

The Newton root-finding iteration for this quadratic polynomial isAt $x_0=1$, it gives $x_1=1/(1-\epsilon )$. Its geometric series expansion is $1+\epsilon +{\epsilon }^2+{\epsilon }^3+O({\epsilon }^4)$. Subtracting the expansion of the polynomial root yields

The definition of hyperbolic sine simplifies the composition exactly for $x>0$:Set $h=x-1$. The geometric series $1/(1+h)=1-h+h^2-h^3+h^4+O(h^5)$, valid for $|h|<1$, then gives

Divide numerator and denominator by $x$ before taking their Maclaurin series:For the reciprocal denominator, the geometric series gives $1-x^2/6+7x^4/360+O(x^6)$. Multiplying gives the Taylor expansionStrictly, the original quotient has a removable singularity at zero; this is the Taylor series of its continuous, analytic extension with value one there. The absence of odd powers also follows because the ratio of these two odd functions is even.

For $x>0$, the definition of hyperbolic cosine simplifies the composite function exactly:Let $h=x-2$. The geometric series givesAdding $2+h$ and dividing by two gives the requested Taylor expansion:The underlying infinite geometric expansion has $|x-2|<2$.

Factor out the dominant cubic term and use a geometric series in $1/(3x^2)$:Thus the requested leading term and absolute error order areThe relative error of retaining only the leading term is $O(x^{-2})$; it differs from the absolute error order $O(x^{-5})$.

Let $a=\pi /3$, $b=2\pi /3$ and take $n$ equal subintervals of width $h=(b-a)/n=\pi /(3n)$. Since sine is continuous, its integral is the limit of the left-endpoint Riemann sumsUsing Euler's formula, the corresponding complex sum is a finite geometric series:Since $(e^{ih}-1)/h\to i$,Taking imaginary parts givesOne can see the same limit entirely within the real sums: the finite trigonometric-sum identity obtained from this geometric sum givesThus the evaluation is explicitly by a Riemann sum, as required.

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 $1/2$.

The absolutely convergent geometric series has sum ${\sum }_{k=0}^{\infty }{r}^k=(1-r{)}^{-1}$ for $0<r<1$. Its Cauchy product with itself isThere are exactly $m+1$ pairs of nonnegative integers whose sum is $m$. Relabeling $j=m+1$ provesAbsolute convergence justifies the multiplication and regrouping; no differentiation of an infinite series is needed.

Normalization of the probability mass function and the geometric series giveThe tail is another geometric series:

Substitute $r=-x^2$ into the geometric series, without taking any derivatives:Thus the coefficient of $x^n$ in the Taylor series is $(-1{)}^{n/2}$ when $n$ is even and zero when $n$ is odd. If nonzero terms are numbered from one, the $n$th term is $(-1{)}^{n-1}{x}^{2n-2}$.

The inner geometric series has common ratio $2/3$, soThe outer geometric series therefore converges as well, with valueAll summands are nonnegative, so there is no cancellation or conditional-convergence issue in evaluating the nested sums.

Let $S_N={\sum }_{n=1}^N(-1{)}^{n+1}/n$. Integrate the finite geometric series identity ${\sum }_{k=0}^{N-1}(-t{)}^k=[1-(-t{)}^N]/(1+t)$ over $0\le t\le 1$:The remainder has absolute value at most ${\int }_0^1{t}^{N}dt=1/(N+1)\to 0$. Hence the partial sums converge, withThis proves both convergence and the value of the alternating harmonic series without substituting an endpoint into an unproved infinite power series. Its convergence is conditional because the corresponding harmonic series of absolute values diverges.