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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.

Expand the right-hand side in the orthonormal basis from part (b). Its coefficients areThus $\mathbf{b}=2\sqrt{2}{\mathbf{e}}_4+\sqrt{3}{\mathbf{e}}_6$. If $\mathbf{x}=c_0{\mathbf{e}}_0+c_4{\mathbf{e}}_4+c_6{\mathbf{e}}_6$, then matrix multiplication gives $A\mathbf{x}=4c_4{\mathbf{e}}_4+6c_6{\mathbf{e}}_6$. Matching coefficients fixes $c_4=\sqrt{2}/2$ and $c_6=\sqrt{3}/6$, while $c_0$ is arbitrary. Consequently all solutions areGeometrically, the solution set is an affine line in a vector space parallel to the kernel of $A$. The zero eigenvalue makes the component in that direction invisible to $A$; the other two components are uniquely determined. The displayed particular solution is perpendicular to the kernel, so it is also the point of this line closest to the origin.

The vector $(1,-1,0{)}^T$ is an eigenvector with eigenvalue $4$. The complementary subspace of vectors with first two components equal is preserved by $A$: a vector $(s,s,t{)}^T$ maps to $(4s+2t,4s+2t,4s+2t{)}^T$. There, the eigenvalue equations reduce to the $2\times 2$ matrix $\left(\begin{array}{cc}4& 2\\ 4& 2\end{array}\right)$, whose characteristic polynomial is $\lambda (\lambda -6)$. For $\lambda =6$, $t=s$; for $\lambda =0$, $t=-2s$.

Normalising by the Euclidean norm, a complete orthonormal basis of eigenvectors isSign changes of any of these eigenvectors are equally valid. The eigenvalues also check part (a): their sum is $10$ and the sum of their squares is $52$.

Adding the diagonal entries gives $\mathrm{tr}A=4+4+2=10$. Expansion along the first row gives $\det A=4(8-4)+2(0-8)=0$. For the remaining trace, direct matrix multiplication givesThus $\mathrm{tr}(A^2)=20+20+12=52$. By multiplicativity of the determinant, $\det (A^2)=(\det A{)}^2=0$. Altogether,

Let $I_n=\int e^{nx}\sin xdx$ and $J_n=\int e^{nx}\cos xdx$. Integrating the trigonometric functions and differentiating $e^{nx}$ in integration by parts gives, modulo constants,Substitution yields $(1+n^2)I_n=e^{nx}(n\sin x-\cos x)$ and $(1+n^2)J_n=e^{nx}(n\cos x+\sin x)$. ThereforeNo division by $n$ was used, so this formula also applies at $n=0$.

Let $F$ be the cumulative distribution function of $X$. Normalisation of the proposed probability density function requires $M={\int }_{-2}^{2}F(y)dy$. The integrated cumulative distribution function on a bounded interval identity givesFor a direct symmetry proof, the continuous symmetric probability distribution satisfies $F(-y)=1-F(y)$. HenceIn particular $g(y)=F(y)/2$ is nonnegative and has total integral one over its stated support.

The expected value is zero, so the variance is $\mathbb{E}[X^2]$. Symmetry givesThe polynomial term contributes $8/3$. Two applications of integration by parts giveso its definite value on $[0,2]$ is $2-10e^{-2}$. Therefore

Since the probability density function is an even function, $xf(x)$ is an odd function. Its integral over the symmetric support vanishes, so

Put $N=6-2e^{-2}$. The function graph is symmetric about the vertical axis, zero outside $[-2,2]$, and has its unique maximum at $x=0$ with height $2/N$. On $0<x<2$, it decreases as $(1+e^{-x})/N$; the negative side is its reflection. The one-sided interior endpoint values are $(1+e^{-2})/N$, so the probability density function jumps to zero at both ends of its support. It also has a corner at its central maximum because of the absolute value in the exponent.

Values assigned exactly at $x=\pm 2$ do not affect any probability.

A probability density function must have total integral one. The numerator is an even function, soThis is positive, ensuring nonnegativity of the probability density function.