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First of all, I'm not a native english speaker and I'm not sure if I should be calling this a series or a sequence, I'm guessing that this is called a sequence in english.

I'm trying to evaluate the convergence of following sequence using D'Alembert theorem.

$$\sum_{n=1}^{\infty} \frac{n}{1+2^n} $$

After applying and rewriting I get the following

$$\lim_{n\rightarrow\infty} \frac{n+1}{1+2^{n+1}} * \frac{1+2^n}{n}$$

According to my textbook the form given here should simplify to 1. However I don't see how this would simplify to 1. $$\lim_{n\rightarrow\infty}\frac{n+1}{n}$$

I tried writing the multiplication out which would give me the following result. Which does not simplify to the same function as the one that I would get if the function given above would equal to 1.

$$\lim_{n\rightarrow\infty}\frac{n*(2^n+1)+1+2^n}{n(1+2^{n+1})}$$

I don't see where I'm going wrong here, any tips are very much appreciated.

  • 0
    $$\lim_{n\rightarrow\infty} \frac{n+1}{1+2^{n+1}} \times \frac{1+2^n}{n}=\lim_{n\rightarrow\infty} \frac{n}{2^{n+1}} \times \frac{2^n}{n}=\frac12$$ – 2017-01-22
  • 0
    Why are you allowed to just drop all the $+1$'s? – 2017-01-22
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    $n+1\sim n$ as $n\to \infty$. In deed $n+1=n\left(\frac{1}{n}+1\right)$ – 2017-01-22
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    In English this test is called the ratio test. – 2017-01-22

1 Answers 1

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$$L=\lim_{n\rightarrow\infty}\left |\frac{a_{n+1}}{a_n}\right|=\lim_{n\rightarrow\infty}\left |\frac{\frac{n+1}{1+2^{n+1}}}{\frac{n}{1+2^{n}}}\right|=\lim_{n\rightarrow\infty}\left |\frac{\frac{n\left(1+\frac 1n\right)}{2^{n+1}\left(\frac{1}{2^{n+1}}+1\right)}}{\frac{n}{2^{n}\left(\frac 1{2^n}+1\right)}}\right|=\frac 12$$

Note $$\lim_{n\rightarrow\infty}\left(1+\frac 1n\right)=\lim_{n\rightarrow\infty}\left(\frac{1}{2^{n+1}}+1\right)=\lim_{n\rightarrow\infty}\left(\frac 1{2^n}+1\right)=1$$