Faktorial bolak-balik menaik

• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots }$. Ini dikenal "deret harmonik bolak-balik".
• ${\displaystyle \ln 2={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)}}}$.
• ${\displaystyle \ln 2={\frac {5}{8}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)}}}$.
• ${\displaystyle \ln 2={\frac {2}{3}}+{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}}$.
• ${\displaystyle \ln 2={\frac {131}{192}}+{\frac {3}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}}}$.
• ${\displaystyle \ln 2={\frac {661}{960}}+{\frac {15}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}}}$.

Faktorial konstanta menaik biner

• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}}$.
• ${\displaystyle \ln 2=1-\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)}}}$.
• ${\displaystyle \ln 2={\frac {1}{2}}+2\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)}}}$.
• ${\displaystyle \ln 2={\frac {5}{6}}-6\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)}}}$
• ${\displaystyle \ln 2={\frac {7}{12}}+24\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)}}}$.
• ${\displaystyle \ln 2={\frac {47}{60}}-120\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)}}}$.

Wakilan deret lainnya

• ${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+2)}}=\ln 2}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(4n^{2}-1)}}=2\ln 2-1}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(9n^{2}-1)}}=2\ln 2-{\frac {3}{2}}}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4n^{2}-2n}}=\ln 2}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {2(-1)^{n+1}(2n-1)+1}{8n^{2}-4n}}=\ln 2}$.
• ${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+1}}={\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}}$.
• ${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+2}}=-{\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}}$.
• ${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2}$ menggunakan ${\displaystyle \lim _{N\rightarrow \infty }\sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4k^{2}-3k}}=\ln 2+{\frac {\pi }{6}}}$ (jumlah timbal-balik dari bilangan dekagonal).

Melibatkan fungsi zeta Riemann

• ${\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}}$.
• ${\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2n+1}}[\zeta (n)-1]=1-\gamma -{\frac {\ln 2}{2}}}$.
• ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{2n-1}(2n+1)}}\zeta (2n)=1-\ln 2}$.

(${\displaystyle \gamma }$ adalah konstanta Euler−Mascheroni dan ${\displaystyle \zeta }$ adalah fungsi zeta Riemann.)

Wakilan tipe-BBP

${\displaystyle \ln 2={\frac {2}{3}}+{\frac {1}{2}}\sum _{k=1}^{\infty }\left({\frac {1}{2k}}+{\frac {1}{4k+1}}+{\frac {1}{8k+4}}+{\frac {1}{16k+12}}\right){\frac {1}{16^{k}}}}$

(Lihat lebih banyak mengenai wakilan tipe Bailey−Borwein−Plouffe (BBP).)

Menerapkan ketiga deret umum untuk logaritma natural ke 2 secara langsung memberikan:

• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}}$.
• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}}$.
• ${\displaystyle \ln 2={\frac {2}{3}}\sum _{k=0}^{\infty }{\frac {1}{9^{k}(2k+1)}}}$.

Menerapkannya untuk ${\displaystyle \textstyle {2={\frac {3}{2}}\cdot {\frac {4}{3}}}}$ memberikan:

• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{2^{n}n}}+\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{3^{n}n}}}$.
• ${\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{3^{n}n}}+\sum _{n=1}^{\infty }{\frac {1}{4^{n}n}}}$.
• ${\displaystyle \ln 2={\frac {2}{5}}\sum _{k=0}^{\infty }{\frac {1}{25^{k}(2k+1)}}+{\frac {2}{7}}\sum _{k=0}^{\infty }{\frac {1}{49^{k}(2k+1)}}}$.

Menerapkannya untuk ${\displaystyle 2=\left({\sqrt {2}}\right)^{2}}$ memberikan:

• ${\displaystyle \ln 2=2\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{({\sqrt {2}}+1)^{n}n}}}$.
• ${\displaystyle \ln 2=2\sum _{n=1}^{\infty }{\frac {1}{(2+{\sqrt {2}})^{n}n}}}$.
• ${\displaystyle \ln 2={\frac {4}{3+2{\sqrt {2}}}}\sum _{k=0}^{\infty }{\frac {1}{(17+12{\sqrt {2}})^{k}(2k+1)}}}$.

Menerapkannya untuk ${\displaystyle \textstyle 2={\left({\frac {16}{15}}\right)}^{7}\cdot {\left({\frac {81}{80}}\right)}^{3}\cdot {\left({\frac {25}{24}}\right)}^{5}}$ memberikan:

• ${\displaystyle \ln 2=7\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{15^{n}n}}+3\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{80^{n}n}}+5\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{24^{n}n}}}$.
• ${\displaystyle \ln 2=7\sum _{n=1}^{\infty }{\frac {1}{16^{n}n}}+3\sum _{n=1}^{\infty }{\frac {1}{81^{n}n}}+5\sum _{n=1}^{\infty }{\frac {1}{25^{n}n}}}$.
• ${\displaystyle \ln 2={\frac {14}{31}}\sum _{k=0}^{\infty }{\frac {1}{961^{k}(2k+1)}}+{\frac {6}{161}}\sum _{k=0}^{\infty }{\frac {1}{25921^{k}(2k+1)}}+{\frac {10}{49}}\sum _{k=0}^{\infty }{\frac {1}{2401^{k}(2k+1)}}}$.

Contoh

Jika ${\displaystyle p^{s}=q^{t}+d}$ dengan beberapa ${\displaystyle d}$, maka ${\displaystyle {\frac {p^{s}}{q^{t}}}=1+{\frac {d}{q^{t}}}}$ dan karena itu

${\displaystyle s\ln(p)-t\ln(q)=\ln \left(1+{\frac {d}{q^{t}}}\right)=\sum _{m=1}^{\infty }(-1)^{m+1}{\frac {({\frac {d}{q^{t}}})^{m}}{m}}=\sum _{n=0}^{\infty }{\frac {2}{2n+1}}{\left({\frac {d}{2q^{t}+d}}\right)}^{2n+1}}$

Memilih ${\displaystyle q=2}$ mewakili ${\displaystyle \ln(p)}$ oleh ${\displaystyle \ln 2}$ dan sebuah deret dari sebuah parameter ${\displaystyle {\frac {d}{q^{t}}}}$ yang ingin tetap kecil untuk konvergen cepat. Mengambil ${\displaystyle 3^{2}=2^{3}+1}$, sebagai contoh, menghasilkan

${\displaystyle 2\ln(3)=3\ln 2-\sum _{k\geq 1}{\frac {(-1)^{k}}{8^{k}k}}=3\ln 2+\sum _{n=0}^{\infty }{\frac {2}{2n+1}}{\left({\frac {1}{2\cdot 8+1}}\right)}^{2n+1}}$:

Ini sebenarnya baris ketiga dalam tabel ekspansi tipe ini:

${\displaystyle s}$	${\displaystyle p}$	${\displaystyle t}$	${\displaystyle q}$	${\displaystyle {\frac {d}{q^{t}}}}$
1	3	1	2	${\displaystyle {\frac {1}{2}}=0.500\,000\,00\dots }$
1	3	2	2	${\displaystyle -{\frac {1}{4}}=-0.250\,000\,00\dots }$
2	3	3	2	${\displaystyle {\frac {1}{8}}=0.125\,000\,00\dots }$
5	3	8	2	${\displaystyle -{\frac {13}{256}}=-0.050\,781\,25\dots }$
12	3	19	2	${\displaystyle {\frac {7153}{524288}}=0.013\,643\,26\dots }$
1	5	2	2	${\displaystyle {\frac {1}{4}}=0.250\,000\,00\dots }$
3	5	7	2	${\displaystyle -{\frac {3}{128}}=-0.023\,437\,50\dots }$
1	7	2	2	${\displaystyle {\frac {3}{4}}=0.750\,000\,00\dots }$
1	7	3	2	${\displaystyle -{\frac {1}{8}}=-0.125\,000\,00\dots }$
5	7	14	2	${\displaystyle {\frac {423}{16384}}=0.025\,817\,87\dots }$
1	1	3	2	${\displaystyle {\frac {3}{8}}=0.375\,000\,00\dots }$
2	11	7	2	${\displaystyle -{\frac {7}{128}}=-0.054\,687\,50\dots }$
11	11	38	2	${\displaystyle {\frac {10433763667}{274877906944}}=0.037\,957\,81\dots }$
1	13	3	2	${\displaystyle {\frac {5}{8}}=0.625\,000\,00\dots }$
1	13	4	2	${\displaystyle -{\frac {3}{16}}=-0.187\,500\,00\dots }$
3	13	11	2	${\displaystyle {\frac {149}{2048}}=0.072\,753\,91\dots }$
7	13	26	2	${\displaystyle -{\frac {4360347}{67108864}}=-0.064\,974\,23\dots }$
10	13	37	2	${\displaystyle {\frac {419538377}{137438953472}}=0.003\,052\,54\dots }$
1	17	4	2	${\displaystyle {\frac {1}{16}}=0.062\,500\,00\dots }$
1	19	4	2	${\displaystyle {\frac {3}{16}}=0.187\,500\,00\dots }$
4	19	17	2	${\displaystyle -{\frac {751}{131072}}=-0.005\,729\,68\dots }$
1	23	4	2	${\displaystyle {\frac {7}{16}}=0.437\,500\,00\dots }$
1	23	5	2	${\displaystyle -{\frac {9}{32}}=-0.281\,250\,00\dots }$
2	23	9	2	${\displaystyle {\frac {17}{512}}=0.033\,203\,12\dots }$
1	29	4	2	${\displaystyle {\frac {13}{16}}=0.812\,500\,00\dots }$
1	29	5	2	${\displaystyle -{\frac {3}{32}}=-0.093\,750\,00\dots }$
7	29	34	2	${\displaystyle {\frac {70007125}{17179869184}}=0.004\,074\,95\dots }$
1	31	5	2	${\displaystyle -{\frac {1}{32}}=-0.031\,250\,00\dots }$
1	37	5	2	${\displaystyle {\frac {5}{32}}=0.156\,250\,00\dots }$
4	37	21	2	${\displaystyle -{\frac {222\,991}{2\,097\,152}}=-0.106\,330\,39\dots }$
5	37	26	2	${\displaystyle {\frac {2\,235\,093}{67\,108\,864}}=0.033\,305\,48\dots }$
1	41	5	2	${\displaystyle {\frac {9}{32}}=0.281\,250\,00\dots }$
2	41	11	2	${\displaystyle -{\frac {367}{2048}}=-0.179\,199\,22\dots }$
3	41	16	2	${\displaystyle {\frac {3385}{65\,536}}=0.051\,651\,00\dots }$
1	43	5	2	${\displaystyle {\frac {11}{32}}=0.343\,750\,00\dots }$
2	43	11	2	${\displaystyle -{\frac {199}{2048}}=-0.097\,167\,97\dots }$
5	43	27	2	${\displaystyle {\frac {12\,790\,715}{134\,217\,728}}=0.095\,298\,25\dots }$
7	43	38	2	${\displaystyle {\frac {3\,059\,295\,837}{274\,877\,906\,944}}=0.011\,129\,65\dots }$

Dimulai dari logaritma natural dari ${\displaystyle q=10}$, salah satunya dapat menggunakan parameter-parameter ini:

${\displaystyle s}$	${\displaystyle p}$	${\displaystyle t}$	${\displaystyle q}$	${\displaystyle {\frac {d}{q^{t}}}}$
10	2	3	10	${\displaystyle {\frac {3}{125}}=0.024\,000\,00\dots }$
21	3	10	10	${\displaystyle {\frac {460\,353\,203}{10\,000\,000\,000}}=0.046\,035\,32\dots }$
3	5	2	10	${\displaystyle {\frac {1}{4}}=0.250\,000\,\dots }$
10	5	7	10	${\displaystyle -{\frac {3}{128}}=-0.023\,437\,50\dots }$
6	7	5	10	${\displaystyle {\frac {17\,649}{100\,000}}=0.176\,490\,00\dots }$
13	7	11	10	${\displaystyle -{\frac {3\,110\,989\,593}{100\,000\,000\,000}}=-0.031\,109\,90\dots }$
1	11	1	10	${\displaystyle {\frac {1}{10}}=0.100\,000\,00\dots }$
1	13	1	10	${\displaystyle {\frac {3}{10}}=0.300\,000\,00\dots }$
8	13	9	10	${\displaystyle -{\frac {184\,269\,279}{1\,000\,000\,000}}=-0.184\,269\,28\dots }$
9	13	10	10	${\displaystyle {\frac {604\,499\,373}{10\,000\,000\,000}}=0.060\,449\,94\dots }$
1	17	1	10	${\displaystyle {\frac {7}{10}}=0.700\,000\,00\dots }$
4	17	5	10	${\displaystyle -{\frac {16\,479}{100\,000}}=-0.164\,790\,00\dots }$
9	17	11	10	${\displaystyle {\frac {18\,587\,876\,497}{100\,000\,000\,000}}=0.185\,878\,76\dots }$
3	19	4	10	${\displaystyle -{\frac {3141}{10\,000}}=-0.314\,100\,00\dots }$
4	19	5	10	${\displaystyle {\frac {30\,321}{100\,000}}=0.303\,210\,00\dots }$
7	19	9	10	${\displaystyle -{\frac {106\,128\,261}{1\,000\,000\,000}}=-0.106\,128\,26\dots }$
2	23	3	10	${\displaystyle -{\frac {471}{1000}}=-0.471\,000\,00\dots }$
3	23	4	10	${\displaystyle {\frac {2167}{10\,000}}=0.216\,700\,00\dots }$
2	29	3	10	${\displaystyle -{\frac {159}{1000}}=-0.159\,000\,00\dots }$
2	31	3	10	${\displaystyle {\frac {39}{1000}}=-0.039\,000\,00\dots }$