# Does the series $\sum\frac{a_n}{a_{n+1}}\frac{1}{n}$ always diverge if $0<a_n<1$?

Does the series $$\sum_{n=1}^{\infty}\frac{a_n}{a_{n+1}}\frac{1}{n}$$ diverge for any $a_n$, satisfying $0<a_n<1$, $n=1, 2, 3\dots$ ?

• It is an interesting problem. Do you know the source? – Davide Giraudo Jun 1 '14 at 15:28
• I came to this problem from another problem of Monthly Journal. – Mher Jun 2 '14 at 12:07
• Maybe you could say it in the body of the question. – Davide Giraudo Jun 2 '14 at 19:11

For each $n$, let $b_n\in (-\infty,0)$ such that $\exp(b_n)=a_n$. By convexity of the exponential function on $\mathbb R$, $$\tag{1}\exp\left(\frac 1M\sum_{j=1}^M(b_j-b_{j+1})\right)\leqslant \frac 1M\sum_{j=1}^M\exp(b_j-b_{j+1})=\frac 1M\sum_{j=1}^M\frac{a_j}{a_{j+1}}.$$ Assume that $\sum_{j=1}^\infty\frac{a_j}{a_{j+1}}\frac 1j$ is convergent. Since for each $N\leqslant M$, $$\frac 1M\sum_{j=1}^M\frac{a_j}{a_{j+1}}\leqslant \frac 1M\sum_{j=1}^N\frac{a_j}{a_{j+1}}+\sum_{j=N}^M\frac{a_j}{a_{j+1}}\underbrace{\frac 1M}_{\leqslant \frac 1j}\leqslant\frac 1M\sum_{j=1}^N\frac{a_j}{a_{j+1}}+\sum_{j=N}^M\frac{a_j}{a_{j+1}}\frac 1j,$$ we obtain that $$\limsup_{M\to +\infty}\frac 1M\sum_{j=1}^M\frac{a_j}{a_{j+1}}\leqslant\sum_{j=N}^\infty\frac{a_j}{a_{j+1}}\frac 1j,$$ hence by (1), $$\lim_{M\to +\infty}\exp\left(\frac 1M\sum_{j=1}^M(b_j-b_{j+1})\right)=0.$$ Since $\sum_{j=1}^M(b_j-b_{j+1})=b_1-b_{M+1}$, it follows that $$\lim_{M\to +\infty}\exp(-b_{M+1}/M)=0$$ which is not possible because each $b_M$ is negative.
The argument would boil down if we had considered $\sum_{n=1}^\infty\frac{a_{n+1}}{na_n}$ and in this case the series may converge, for example with $a_n:=2^{-n^2}$.
{"ft_lang_label":"__label__en","ft_lang_prob":0.53725857,"math_prob":0.99999905,"size":1671,"snap":"2019-51-2020-05","text_gpt3_token_len":753,"char_repetition_ratio":0.2015597,"word_repetition_ratio":0.011904762,"special_character_ratio":0.42309994,"punctuation_ratio":0.07368421,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000098,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-11T14:49:37Z\",\"WARC-Record-ID\":\"<urn:uuid:8020a9a3-a142-400b-a004-181423445797>\",\"Content-Length\":\"135870\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:82700ddb-f1a1-4819-9457-63e509a34884>\",\"WARC-Concurrent-To\":\"<urn:uuid:0d08ba6c-ce7e-44ac-bb5d-776ab69dc2bd>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/816920/does-the-series-sum-fraca-na-n1-frac1n-always-diverge-if-0a-n1/817015\",\"WARC-Payload-Digest\":\"sha1:SXLJHSXU7WB6P34TFN6U7SYU5CTZ26VL\",\"WARC-Block-Digest\":\"sha1:G3OLZH6RSKTCHMNDPDD5CQBNWZJTAJEP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540531917.10_warc_CC-MAIN-20191211131640-20191211155640-00555.warc.gz\"}"}