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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.

Evaluate the absolute values on the three intervals separated by one and two:The strict inequality therefore holds precisely for

Write $\mathbf{\Delta }=\mathbf{d}-\mathbf{b}$. At an intersection, $\mathbf{\Delta }=\lambda \mathbf{a}-\mu \mathbf{c}$, so its scalar triple product with $\mathbf{a},\mathbf{c}$ vanishes. The expression in the question is exactlyUnder a shift of origin by $\mathbf{h}$, the direction vectors $\mathbf{a},\mathbf{c}$ stay fixed, while $\mathbf{b},\mathbf{d}$ both change by $-\mathbf{h}$. Thus $\mathbf{\Delta }$ and the condition are invariant. For nonparallel lines this is also sufficient for intersection; for parallel lines it only asserts coplanarity and need not imply intersection.

Put $\delta \lambda ={\lambda }_2-{\lambda }_1$ and $\delta \mu ={\mu }_2-{\mu }_1$. Then ${\mathbf{x}}_2-{\mathbf{x}}_1=\delta \lambda \mathbf{a}$ and ${\mathbf{x}}_4-{\mathbf{x}}_1=({\mathbf{x}}_3-{\mathbf{x}}_1)+\delta \mu \mathbf{c}$. Bilinearity of the cross product givesThe tetrahedron volume is one sixth of the absolute value of this triple product. Sliding either vertex pair along its line preserves $|\delta \lambda |$ or $|\delta \mu |$, because its separation is fixed and the direction vectors are fixed. Thus the volume is unchanged, including the zero-volume coplanar case. This is tetrahedron volume under sliding opposite edges.

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

The line has equation $2x-y+2=0$, with normal vector $(2,-1)$. The distance from a point to a line is the absolute value of the normal component of the displacement:The perpendicular foot is $(4,-6)-(16/5)(2,-1)=(-12/5,-14/5)$, which lies on the line and realizes this shortest distance.

The numerator is the derivative of the denominator. Using the chain rule in reverse gives the antiderivativeThe absolute value makes this real on every interval where the denominator is nonzero. The integration constant may be chosen independently on the intervals separated by $x=(-3\pm \sqrt{17})/2$.

Differentiate $x=r\cos \theta$, $y=r\sin \theta$:Squaring and adding cancels the mixed terms and uses ${\cos }^2\theta +{\sin }^2\theta =1$, givingFor this cardioid, the polar arc length integrand isThe cosine inside the absolute value changes sign at $\theta =\pi$. Therefore the perimeter isin coordinate length units. The isolated cusp does not obstruct this piecewise smooth arc length integral.

As $x\to 0$ through nonzero values, $1/x^2\to +\infty$ and $e^{-1/x^2}\to 0=f(0)$. Thus $f$ is continuous at zero.

For the derivative at zero use its definition, rather than substituting into a formula obtained away from zero:Indeed, putting $t=1/|h|$ makes the absolute value $t{e}^{-t^2}$, which tends to zero since Gaussian decay dominates every power. Therefore $f$ is continuous and differentiable at zero, with $f^{\prime }(0)=0$. Away from zero, the chain rule gives $f^{\prime }(x)=2x^{-3}{e}^{-1/x^2}$.

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

The absolute value in the original PDF reflects every negative sine arch upward. The graph is nonnegative, with zeros at $x=k\pi$ for $k=0,1,2,3,4$, and it lies below the decaying envelope $e^{-x}$. On the $k$th arch, differentiation gives a factor proportional to $\cos x-\sin x$. Thus its unique local maximum isSuccessive maximum heights have ratio $e^{-\pi }$. At each interior zero the one-sided slopes have opposite signs, producing a corner. The smaller panels below magnify the later arches so that all four remain visible.

For $0<x<1$,In particular its absolute value is bounded by $|b|x^{\min (b,0)}$, proving convergence when $b>-1$; the apparent endpoint singularity at one is removable, with value $b$. On any compact parameter interval with lower bound $b_0>-1$, the differentiated integrand is $x^b$, dominated by $x^{b_0}$. The dominated convergence theorem therefore justifies differentiation under the integral sign:Hence $J(b)=\ln (b+1)+C$. Since the integrand is identically zero at $b=0$, $J(0)=0$ fixes $C=0$. ThusThis proves the logarithmic integral representation of the natural logarithm, including its parameter range.

The square roots simplify to the absolute values $|a-b|$ and $|a+b|$. Put $s=\sin (\pi \sqrt{a^2+b^2})$. A nontrivial solution of the homogeneous linear system exists exactly when its coefficient matrix is singular:The second factor vanishes precisely when $(a-b{)}^2=(a+b{)}^2$, or $ab=0$, giving both coordinate axes. The first factor vanishes when $\sqrt{a^2+b^2}$ is a nonnegative integer. Hence the complete locus isThese are the two axes together with concentric circles of positive integer radii. The zero-radius case is the origin, already included on the axes; at that point both equations vanish and any nonzero pair $(x,y)$ works.

The absolute value is nonnegative, so a real point must satisfy $1-x^2\ge 0$, or $-1\le x\le 1$. On that interval the two branches areThey form a closed lens symmetric about both coordinate axes, meeting at $(\pm 1,0)$ and reaching $(0,\pm 1)$. The parabolas outside the stated interval do not belong to the original locus.

Substituting $y=1/2$ into the absolute value relation gives $1/2=1-x^2$, so $x^2=1/2$. All real intersections are

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.

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.

First suppose $\alpha >0$. Differentiation under the integral sign, justified on compact positive parameter intervals by an integrable bound, givesThe first integral is $f(\alpha )/\alpha$. For $\alpha \ne 1$, use the partial fraction decompositionThe standard inverse tangent antiderivatives then giveThe same value holds at $\alpha =1$ by continuity, or by direct integration. Hence the proposed first-order linear differential equation is valid on the positive parameter range:Divide by $\alpha$ to recognise a quotient rule derivative:To fix $C$, the value $f(0)=0$ alone is insufficient, since every displayed solution tends to zero there. Instead considerThis limit follows from the dominated convergence theorem: for $0<\alpha \le 1$ and $0<x<1$, the logarithm's absolute value is bounded by $2|\ln x|+\ln 2$; for $x\ge 1$, it is at most $\ln (1+x^2)$. After division by $1+x^2$, these are integrable. The limiting improper integral is absolutely convergent, and the substitution rule $x=1/t$ changes it into its own negative. Its value is therefore zero, giving $C=0$. ThusThe question calls $\alpha$ a real parameter, but this last formula and the proposed differential equation do not hold on the entire real axis. The defining integral is odd in $\alpha$, because its logarithm depends on ${\alpha }^2$. Applying the positive-parameter result to $|\alpha |$ therefore gives the correct extensionwith $f(0)=0$. In particular, for $\alpha <0$, the right-hand side of the differential equation must be $-\pi \alpha /(1-\alpha )$; for every $\alpha \ne 0$ the unified expression is $\pi |\alpha |/(1+|\alpha |)$. The logarithmic quadratic parameter integral therefore provides the requested identity on its valid range and the corrected answer elsewhere.

Completing the square under the square root gives $\sqrt{(x-1{)}^2}=|x-1|$. Thus the equation is $|x-1|=2$, and the definition of absolute value gives $x-1=\pm 2$. Hence

Put $r=\sin (\pi /4)=1/\sqrt{2}$. The geometric series has first term $r$ and common ratio $-r$, with absolute value less than one. Therefore its sum is

For $x>0$, the denominator is positive, so the function graph has the same sign as $\sin x$. Its crossings of the horizontal axis areEach crossing is genuine because the derivative there is $(-1{)}^n/(n\pi {)}^2\ne 0$. Near zero, $\sin x\sim x$, so $y\sim 1/x\to +\infty$; the origin is excluded and is not a crossing. The oscillations alternate sign and decay towards zero as $x\to \infty$, with absolute value at most $1/x^2$.