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5

Using l'Hopital's rule, we find $$ \lim_{x \to 1} \frac {\ln^2 x}{1-x} = \lim_{x \to 1} \frac {2 \ln x \cdot \frac {1}{x}}{-1} = \lim_{x \to 1} - \frac {2 \ln x}{x} = \lim_{x \to 1} 0 = \boxed {0}. $$We know l'Hopital's rule applies because it is of the form $ \frac {\ln^2 1}{1-1} = \frac {0}{0} $, which is indeed indeterminate.


5

The line you write is incorrect, $1+p+\ldots+p^{n-1}$ approaches to $\frac1{1-p}$ assuming that $|p|<1$. Namely, if $|p|<1$, then $\lim\limits_{n\to\infty}(1+\ldots+p^n)=\dfrac1{1-p}$. This is true exactly because $1+\ldots+p^{n-1}=\frac{1-p^n}{1-p}$, and when $|p|<1$, taking $n$ to infinity yields $p^n\to 0$.


4

We can use approximations : $\ln(x) \approx x-1$ when $x$ is close to $1$. Hence the required limit reduces to $\lim_{x \to 1}(1-x)=0$


4

Hint: $$n!>\left(\frac{n}{2}\right )^n$$ for large enough $n$.


4

Pick $K$ as large as you want, then $n!\ge K^{n-K}$ (for $n\ge K$), so $$(n!)^{1/n}\ge K^{(n-K)/n}=K^{1-K/n}\to K\text{ as }n\to\infty.$$


4

It's easier to shift indices, so that the summand looks the same between the sums you are comparing. Your sum is $$\sum_{k=2}^{n+1} \frac{t^k}{k!}.$$ You can now "add and subtract" the first two terms of the sum for the exponential: $$\sum_{k=2}^{n+1} \frac{t^k}{k!} = \sum_{k=0}^{n+1} \frac{t^k}{k!} - \sum_{k=0}^1 \frac{t^k}{k!} \\ = \sum_{k=0}^{n+1} ...


3

$(2r+1)r = 2(r-1)r + 3r = 4\binom{r}{2} + 3\binom{r}{1}$


3

Let $s_n$ and $t_n$ denote the $n$th partial sums of $\sum_{n = 0}^\infty a_{2n}$ and $\sum_{n = 0}^\infty a_{2n+1}$, respectively. If $r_n$ is the $n$th partial sum of $\sum_{n=0}^\infty a_n$, then $r_{2n} = s_n + t_{n-1}$ and $r_{2n-1} = s_{n-1} + t_{n-1}$, for all $n \ge 1$. Since $\sum_{n = 0}^\infty a_{2n}$ and $\sum_{n = 0}^\infty a_{2n+1}$ are ...


3

Here are the steps $$ \lim\limits_{x\to 1} \frac{(\ln x)^2}{1-x} = \lim\limits_{x\to 1} \frac{\frac{d}{dx}(\ln x)^2}{\frac{d}{dx}[1-x]} $$ $$ =\lim\limits_{x\to 1} \frac{2(\ln x)\frac{d}{dx}[\ln x]}{\frac{d}{dx}[1]-\frac{d}{dx}[x]} =2\lim\limits_{x\to 1} \frac{\frac{\ln x}{x}}{0-1} $$ $$ =-2\lim\limits_{x\to 1} \frac{\ln x}{x} ...


3

As the number of terms of the Arithmetic sequence is odd, let the terms be $a,a\pm d,a\pm 2d$ So, $a-d+a-2d+a+a+d+a+2d=5\implies a=1$ and $(a-d)(a-2d)(a)(a+d)(a+2d)=1155$ $$\iff(a^2-d^2)(a^2-4d^2)=1155$$ Had the number of terms of the Arithmetic sequence been even, we could take the terms to be $a\pm d,a\pm3d,a\pm5d$ etc.


3

This is a quick and slightly unorthodox approach. By inspection, $$1155=1\times 3\times 5\times 7\times 11$$ The five numbers would have formed a nice AP apart from the missing $9$. Also, they sum to $27$, which is $12$ in excess of the required $15$. Let's try and even out the spacing by making some of the numbers negative. As the excess of $12$ ...


2

by L'Hospital we get $$\lim_{x \to 1}2\ln(x)\frac{1}{x}(-1)=0$$


2

So, the first summation indexes from $1$ to $m^2-1$. The next line simply breaks that summation into chunks. The left Summation indexes from $1$ to $m-1$ while the second summation will take that index and sum up to one less than the next number squared. So, instead of one summation indexing $1,2,3,4,\dots m^2-1$, it is multiple summations. $$\Sigma_1^3 ...


2

I think you have the right idea, but you are right to be concerned about the fact that in your equation (1), the sum always starts at an even index and ends at an odd index, so it's not quite general. Instead of starting with $$\left|\sum_{k=n}^{m}({a_{2k}+a_{2k+1}})\right|$$ (which always starts at the even index $2n$ and ends at the odd index $2m+1$), ...


2

Note that $x(n+1-x)\ge (1)(n)$ for $1\le x\le n$. It follows that $$(n!)^2=(1)(n)(2)(n-1)(3)(n-2)\cdots (n)(1)\ge n^n$$ and therefore $(n!)^{2/n}\ge n$.


2

Do you know $\sum_{r=1}^n r^2=\dfrac{n(n+1)(2n+1)}6$ and $\sum_{r=1}^n r=\dfrac{n(n+1)}2$ also for integer $n,r\ge0,$ $$\binom nr=\frac{n(n-1)\cdots(n-r+1)}{r!}$$ consequently, $\binom n2=\dfrac{n(n-1)}2$ and $\binom n1=n$


2

$e^{t}=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+ \cdots$, then $$e^t-1-t=\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots=\sum_{2}^{\infty}\frac{t^k}{k!}$$


1

how does this work for other examples Not via a Taylor series, that's for sure. The difference between $f$ is the limit of a uniformly converging power series, and $f$ is the limit of a uniformly converging sequence of polynomials is that partial sums of a power series are a very special kind of a sequence of polynomials: for every $k$, the ...


1

Write out the first 5 or so terms of the series for the exponential and the series you have here. What you'll find is that the series you have here is exactly the series for the exponential but starting at the $t^2$ term. Thus if you subtract the first two terms from the exponential you get the series that you have here. To transform one series into the ...


1

Fix $n\in \Bbb N$. Set $\delta = \varepsilon(k_n - 1)$. Then $\delta \ge 0$. Hence, by definition of $k_{n+1}$, there exists a $k\in \Bbb N$ with $k \le (1 +\varepsilon)^{n+1}$ such that $k_{n+1} \le k + \delta$. Thus $$k_{n+1} \le (1 + \varepsilon)^{n+1} + \varepsilon(k_n - 1) < (1 + \varepsilon)(k_n + 1) + \varepsilon(k_n - 1) = (1 + 2\varepsilon)k_n + ...


1

This is a consequence of the fact that the binomial coefficients form a basis for the space of polynomials; see http://en.wikipedia.org/wiki/Binomial_coefficient#Binomial_coefficients_as_a_basis_for_the_space_of_polynomials . In this case in particular, the easiest way to derive the result is likely going coefficient by coefficient. In general, ${r\choose ...


1

$$\begin{align}\binom r1 =r\qquad &\Rightarrow r=\binom r1\\ \binom r2=\frac {r(r-1)}2\qquad &\Rightarrow r^2=2\binom r2+\binom r1 \end{align}$$ Hence $$\begin{align} \sum_{r=1}^{m-1}(2r+1)r&=\sum_{r=1}^{m-1}2\left[2\binom r2+\binom r1\right]+\binom r1\\ &=\sum_{r=1}^{m-1}4\binom r2+ 3\binom r1\qquad \blacksquare \\ &=4\binom m{3}+3\binom ...


1

Let $c_n=(n!)^2$; then $\displaystyle\frac{c_{n+1}}{c_n}=\frac{((n+1)!)^2}{(n!)^2}=(n+1)^2\to\infty,$ $\;\;\;$so $(n!)^{\frac{2}{n}}=(c_n)^{\frac{1}{n}}\to\infty$


1

Hint. For a) $\displaystyle |a_n\sin a_n|\le|a_n|$ implies $\displaystyle \sum a_n\sin a_n$ is absolutely convergent, thus is convergent. For b), as you correctly found, we have $$ \sqrt{a_n} \: \frac{n^{a_n}-1}{\ln{n}} \sim a_n^{3/2}. \tag1$$ Now, as $n$ is sufficiently great, since $a_n$ tends to $0$, we have $a_n^{3/2}\leq C a_n$ for some constant ...


1

Another approach you could use would be to use the sequences of partial sums: If $\displaystyle T_n=\sum_{k=0}^na_{2k}, \;U_n=\sum_{k=0}^na_{2k+1}, \; \text{and}\;S_n=\sum_{k=0}^na_k$, then $T_n\to T \text{ and }U_n\to U$ for some $T,U\in\mathbb{R}$; so $S_{2n}=T_n+U_{n-1}\to T+U,\;\;\; S_{2n+1}=T_n+U_n\to T+U$, and therefore $S_n\to T+U$.


1

OEIS tells us this is the "Binary partition function: number of partitions of n into powers of 2."


1

Since $\pi\csc(\pi z)$ has residue $(-1)^n$ at $z=n$ for $n\in\mathbb{Z}$, we will use the contours $$ \gamma_\infty=\lim\limits_{R\to\infty}Re^{2\pi i[0,1]}\qquad\text{and}\qquad\gamma_0=\lim\limits_{R\to0}Re^{2\pi i[0,1]} $$ To sum over all $n\in\mathbb{Z}$ except $n=0$, we use the difference of the contours, which circles the non-zero integers once ...


1

No, that's only for ordinary series. You can see by the root test that $$\lim_{n\to\infty}\left|\sqrt{n}z^n\right|^{1/n}=\lim_{n\to\infty}n^{1/2n}|z|=|z|$$ converges absolutely when this limit is $<1$, i.e. when $|z|<1$.


1

Hint Use the ratio test to prove that the radius of convergence is $R=1$. Use the Dirichlet's test to prove the convergence for $|z|=1$ and $z\ne1$, and for $z=1$ use the Leibniz criterion to prove the convergence.


1

An approach could be $$ (1-p)(1+p+p^2+\dots+p^{n-1})=(1+p+p^2+\dots+p^{n-1})-(p+p^2+\dots+p^{n-1}+p^n)=1-p^n $$ giving $$ 1+p+p^2+\dots+p^{n-1}=\frac{1-p^{n}}{1-p}, \quad p\neq1, $$ moreover if $|p|<1$, then $p^n \rightarrow 0$ and you get $$ 1+p+p^2+\dots+p^{n-1}+\ldots=\frac{1}{1-p},\quad |p|<1. $$



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