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The comparison test for series says that $0\le v_n\le C{w}_n$ eventually and convergence of $\sum w_n$ imply convergence of $\sum v_n$. Conversely, an eventual positive lower comparison with a divergent positive series proves divergence. The ratio test uses $L={\lim }_{n\to \infty }{v}_{n+1}/v_n$: $L<1$ proves convergence, $L>1$ proves divergence, and $L=1$ is inconclusive. Removing finitely many initial zero terms does not affect these conclusions.

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 the positive summand $a_n=(n!{)}^3{e}^{3n}/(3n)!$, the ratio test givesThus the series converges absolutely.

Here $|a_{n+1}/a_n|=n/(n+1)\to 1$. Therefore the ratio test is inconclusive. The series does in fact converge by the alternating series test, and fails to converge absolutely because the harmonic series diverges; these facts use other tests.

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 $a_n=n^n/n!$, successive terms satisfyThey eventually grow by a factor strictly greater than one and cannot tend to zero. Hence the series diverges to $+\infty$, by the ratio test or the term test for divergence.

For $a_n=n^2/n!$, the ratio test givesThus the series converges absolutely. As a check, $n^2=n(n-1)+n$ and the exponential power series give the sum $e+e=2e$.

Here $0<\ln 2<1$, so $(\ln 2{)}^{-n}$ grows exponentially rather than decaying. The positive terms $a_n=n^3/(\ln 2{)}^n$ do not tend to zero; indeed they tend to $+\infty$. Therefore the series diverges by the necessary condition that the terms of a convergent series tend to zero.

The ratio test gives the same conclusion: $a_{n+1}/a_n=((n+1)/n{)}^3/\ln 2\to 1/\ln 2>1$.

Let $a_n=2^n/n!$. The ratio test gives $a_{n+1}/a_n=2/(n+1)\to 0<1$, so the series converges absolutely. In fact the exponential series identifies its value aswith the $n=0$ term subtracted.

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.

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 $N\to \infty$.

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 $1/3$.

For the positive terms $a_n=n!/2^n$, the ratio is $a_{n+1}/a_n=(n+1)/2\to \infty$. The ratio test therefore gives divergence. More directly, the terms increase by a factor at least two for all $n\ge 3$, so they fail the necessary term test for divergence condition $a_n\to 0$. Hence

For the positive terms $u_n=2^n/n!$, the consecutive ratio is $u_{n+1}/u_n=2/(n+1)\to 0$, so the ratio test proves convergence. The exponential series gives $e^2={\sum }_{n=0}^{\infty }{2}^n/n!$. Removing the $n=0$ term yields

For a series $\sum u_k$ with eventually nonzero terms, supposeexists, allowing $L=+\infty$. The ratio test says that if $L<1$, the series converges absolutely; if $L>1$, it diverges; if $L=1$, the test is inconclusive. In the divergent case the terms fail to tend to zero. For example, the P-series with exponents $1$ and $2$ both have ratio limit $1$ but have different convergence behaviour.

For $u_k=a^{2k+1}/(2k+1)>0$, the consecutive ratio isThus the ratio test proves convergence when $0<a<1$ and divergence when $a>1$. At $a=1$, the series is ${\sum }_{k\ge 0}1/(2k+1)$, which diverges by comparison with $\frac{1}{2}{\sum }_{k\ge 0}1/(k+1)$. ThereforeAll terms are nonnegative, so convergence here is also absolute convergence.

For the positive terms $a_n=n^4/3^n$, the consecutive ratio isThe ratio test therefore proves that the series converges absolutely. More explicitly, for all sufficiently large $n$ the ratio is at most a fixed $q<1$, so the tail is bounded by a convergent geometric series.

Classification and existence. For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, an ordinary point $x_0$ is one where $p,q$ are analytic. There are two independent analytic power series solutions, with arbitrary initial values $y(x_0),y^{\prime }(x_0)$. A regular singular point is a singular point where $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ are analytic. The Frobenius method then constructs solutions beginning with $(x-x_0{)}^{\sigma }$, where $\sigma$ solves the indicial equation. At least one Frobenius solution exists; two such solutions exist without logarithms when the two exponents differ by a noninteger. Repeated exponents or an integer difference can require a logarithmic second solution.

For this equation,Both are analytic at zero, while $p,q$ themselves have poles there. Thus zero is a regular singular point.

Indicial equation and recurrence. Multiply by $4x$ and insert $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$:The lowest power $x^{\sigma -1}$ has coefficient $[4\sigma (\sigma -1)+2\sigma ]a_0$, givingThe coefficient of $x^{n+\sigma }$ isFor either exponent the common factor is nonzero for $n\ge 0$, so the recurrence relation reduces toNormalizing each leading coefficient to one, the two Frobenius solutions areMore explicitly, $y_2=\sqrt{x}{\sum }_{n\ge 0}{2}^{n}n!x^n/(2n+1)!$. The distinct leading powers establish linear independence, so $y=C_1{y}_1+C_2{y}_2$ on the positive real side of the singular point. The displayed exponential function is the requested closed-form solution; substitution gives $y_1^{\prime }=y_1/2,y_1^{\prime \prime }=y_1/4$, which makes the differential expression zero.

Convergence. For either exponent the ratio of successive terms in the coefficient power series isThus both coefficient series have $\boxed{R=\infty }$ by the ratio test. This does not remove the $\sqrt{x}$ branch point of $y_2$: its analytic coefficient factor is entire, while the full Frobenius solution requires a square-root branch away from zero in the complex plane. The differential equation has no other finite singular points to limit continuation of these coefficient series.

Write a nonterminating parity series as $y=x^r{\sum }_{j\ge 0}{b}_j{x}^{2j}$, where $r=0$ or $1$. Its recurrence relation isThe ratio test applied to its nonzero successive terms has limiting absolute ratio $|x{|}^2$. Hence the series converges absolutely for $|x|<1$ and diverges for $|x|>1$, withAt $x=\pm 1$, the limiting ratio is one, so the basic ratio test alone is inconclusive. Retain the next term of the ratio instead:For large $j$ the ratio is positive. Taking logarithms and summing gives $\log |b_j|=-\frac{3}{2}\log j+O(1)$; the $O(j^{-2})$ remainders have a finite sum. Thus $|b_j|\le C{j}^{-3/2}$ eventually. Comparison with the convergent p-series proves absolute convergence at both endpoints as well:The radius of convergence one is the distance from the expansion point to the nearest singular points of a differential equation, $x=\pm 1$. Convergence of function values at those points does not make the solutions analytic there; endpoint derivatives can diverge. The terminating Chebyshev polynomial is an exception with infinite radius of convergence. For example, at $m=0$ the other solution is $\arcsin x$, whose endpoint values are finite while its first derivative diverges.